There are quite a lot of people who say they can't do maths. I think much of this is down to poor teaching. Your teaching is brilliant. The world needs more maths teachers like you. Never stop teaching!!!!!
Bang on! Maths was always my strongest subject but I saw so many that came to a point where they didn't understand something - usually something quite straightforward - and because it wasn't properly explained to them in language they could comprehend everything after that became a struggle and they effectively gave up. So sad because maths is such an elegant subject.
N is 72 If 2n is a perfect square, then N has an odd power of 2 as a factor. If 3n is a perfect cube, then N has 2^(3A) as a factor and 3^(3B - 1) as a factor. Putting A & B = 1, we get 2^3 * 3^2 = 72. 72*2=144=12^2 72*3=216=6^3
Your solution is very elegant. I found the answer by thinking about prime factors, in a square the prime factors appear in pairs and in a cube they appear in triplets. The value must then have some number of prime factors as 2 and some as three. We can see that there must be an even amount of 3s and an odd amount of 2s. If we look at having only a single 2 as a prime factor and two 3s then it would be a perfect square when multiplied by 2 but not a perfect cube when multiplied by 3. If we instead have three 2s and two 3s as prime factors then it fulfil both conditions. n = 2x2x2x3x3 = 72 2n = (2x2)(2x2)(3x3) = 144 3n = (2x2x2)(3x3x3) = 216
I found the solution using the prime factors..,put n=2^x*3^y.. then 2n=2^(x+1)*3^y.. therefore x is odd .. if we consider 3n, we have 3n=2^x*3^(y+1)..,therefore x must be multiple of 3..the smallest odd number which is multiple of 3 is 3.. then x=3.. furthermore y must be even because 2n is a square.. and y+1 must be a multiple of 3 because 3n is a cube.. the smallest number with these 2 ptoperties is 2..,therefore y=2.. finally we obtain n= 2^3*3^2=72
2n = a² 3n = b³ where a, b and n are positive integers. From here we get 6n = 3a² 6n = 2b³ This leads to the equation 3a² = 2b³ ⟹ a² = 2b³/3 ⟹ a = √(2b³/3). Here obviously 2b³/3 is an integer meaning b has to be divisible by 3: b = 3c ⟹ 2b³/3 = 2*3*3*c³ where c is positive integer. 2*3*3*c³ has to be a perfect square. This mean that the prime factorisation must have even copies of every prime factor. The smallest c to make that happen is c = 2 ⟹ 2*3*3*c³ = 2*2*2*2*3*3 = 12² = a² = 2n ⟹ n = 12²/2 = 72.
Once found n=72, one can generate other numbers satisfying the conditions of 2n being perfect square and 3n being perfect cube by multiplying by p⁶, with p being a positive integer, since that factor is both a perfect square and a perfect cube. Thus, the possible solutions are of the form n=72p⁶, with p=1 being of course the smallest. The next solution is n=4608 (with p=2) 💁🏻♂️
It made complete sense to me. A number is literally used as a given parameter here. In the same fashion you'd argue that the smallest pair (n,m) where 5*n = m^2 should be n=5=m if n and m are ints and greater then zero. Because it literally implies the 'other' n to also be 5 to become a perfect square. Learning this number theory feels so fundamental. I am excited to explore it.
For 2n to be a perfect square n has to be even - divisible by 2. For 3n to be a perfect cube n has to be divisible by 9 and, since it’s even and 3 is not, by 8 as well. The smallest natural number that is divisible by both 8 and 9 is 8x9 = 72 - and it fits .
try 18 2x18 = 36 = 6² 3x18 = 64 = 4³ it must satisfy 2x2 it must satisfy 3x3x3 since 2 and 3 is already the coefficient of n, so it only has to satisfy 2 and 3x3 so 18 it is
Intuitively, you take 2 and 3 (the factors of n), you cube one and square the other, multiply one by the other: 8*9=72. It works because 2 and 3 are primes, and I'm pretty sure it works with other pairs of primes (not too sure about the powers involved, though, you probably need to jump into logs to prove it).
A more elegant solution is to consider the factorization of n. 1) Since 2n is a square and 3n is a cube, the minimal n can only include factors of 2 and 3. So n=2^A*3^B. 2) Since 2n is a square, A must be odd and B even. 3) Since 3n is a cube, A must be a multiple of 3 and B a multiple of three minus one. 4) A is the smallest odd multiple of 3, that is 3 itself. B is the smallest even number that is one less than a multiple of 3, that is 2. 5) So n=2^3*3^2 = 72
Love your videos man, especially those dealing with numbers and number theory. Keep up this wonderful work of making math interesting and accessible. ❤❤❤
A general solution would be: if an = perfect square, bn = pefect cube solution: n=ak^2 abk^2=q^3 k=ab n=ak^2=a*a*a*b*b=a^3*b^2 The lowest integar value of n will always be n = a^3*b^2
Now, the big question would be... is there a value of n so that 3*(n^5) is a perfect cube and 5*(n^3) is a perfect power of 5, and if it's the case, how could you write and prove a generalisation?
I solved by just looking at properties of n. - Has only factors of 2 and 3, otherwise it would be unnecessarily bigger. - Has a number of factors of 2 divisible by 3 (a factor of 8) and a number of factors of 3 divisible by 2 (a factor of 9). - The count of factors of 2 is one less than a multiple of 2. The count of factors of 3 is one less than a multiple of 3. 8 and 9 satisfy this constraint already. Thus multiplying by 2 gives a perfect square, and multiplying by 3 gives a perfect cube. Any number meeting these conditions is sufficient, and the smallest such number is 8 * 9 = 72.
I've been watching a fair number of your videos and I thoroughly enjoy your way of breaking things down. For someone whose interest in pure mathematics is far greater than current skill level, where would you suggest the person should start? Any introduction to pure mathematics you would recommend?
Once you get to 6k^2 =q^3 you can strengthen your immediate jump to k=6 by pointing out that 6 is square-free and so there are no squares or cubes lurking in the factorization of 6.
2n is a perfect square => n is even 3n is a perfect cube mean => a) n contains at least the cube of 2 which is 8 b) n contains at least the square of 3 which is 9 Smallest possible value of n is 8*9=72
There is a very intuitive method which requires no algebra: Consider the characteristics of perfect squares and cubes. A perfect square has only prime factors that come in pairs (such as 2x2) while a perfect cubes has only prime factors which come in triplets (2x2x2). When we add a single 2 to the prime factorization of "n", we get a perfect square, so "n" must already have at least one 2 for that new 2 to pair with. Likewise, n must have at least two 3s. Okay, we now know that "n" has 2s and 3s in its factorization, and if we add a single 2, all the factors will be in pairs, and if we add a single 3, all the factors will be in triplets. Effectively that means that the number of 2s is a multiple of 3 which is one less that a multiple of 2, while the number of 3s is a multiple of 2 which is one less then a multiple of 3. Three 2s and two 3s meets these conditions, and if we starts removing any pairs or triplets of factors then we'll either run out of 2s or 3s, and we've already established that "n" has 2s and 3s in it, so this is also the smallest factorization which will work, and 2x2x2x3x3 is 72. And honestly, this is all belabouring the point quite a bit. Once you start thinking about the prime factorizations and how squares and cubes work, the answer just kind of falls out.
Consider the prime factorization of n. If n has a factor of any prime other than 2 or 3, then we could make a smaller number by taking out all factors of that prime. So, the only prime factors are 2 and 3. So, we define a and b such that n = (2^a) (3^b). Note that 2n has to be a perfect square. So 2^(a+1) 3^b is a perfect square, and hence a+1 and b are both even. If a+1 is even, then a is odd. Furthermore, note that 3n has to be a perfect cube. So 2^a 3^(b+1) is a perfect cube and hence a and b+1 are both multiples of 3. If b+1 is a multiple of 3, then b = 3k+2 for some integer k. The smallest odd multiple of 3 is 3, so a is 3. The smallest even number of the form 3k+2 is 2, so b is 2. Thus, the smallest value of n is (2^3) (3^2) = (8)(9) = 72
I knew that initial insight, then I just grew the sequences starting with 3 x 2 for the 2n is square, and 2 x 3 for the 3n is cubed. Grown in tandem. Easy.
@@chaosredefined3834 Yes, very nice as always. My method was very sloppy, and to explain how and why it would work to someone else would have got me into tangles.
to be honest i wasnt able to understand the first notion of 4k² = 2n , BUT what i understood based on the statement was 2 times n is a perfect square, which means if the perfect square itself was being square root it should be equal to 2 times n , 2.n = square root of perfect square variable , that being said the perfect square variable is equal to (2.n)^2 , and we can substitute the variable with letter x = 4.n^2
If 2n=i^2 and 3n=j^3, we can deduce that ‘i’ can only be an even value due to ‘n’ being restricted to integers and being multiplied by an even integer. We can also deduce that ‘j’ must be a multiple of 3 for the same reasoning. From the given equations, we also know i^2/2=j^3/3=n. After this, we could check every multiple of 6 to replace ‘i’ or ‘j’ and solve for the other, and I would recommend testing inputs for ‘j’ since it has the potential to grow larger results faster. It just so happens j=6 results in a i=12 & n=72, which since ‘i’ and ‘j’ are integers, they create their respective square and cube.
I did it a slightly different way by figuring out that the prime factorization of n would require: - an odd number of 2's and an even number of 3's, in order for 2n to be a perfect square - a number of 3's congruent to 2 mod 3, and a number of 2's that was a multiple of 3, in order for 3n to be a perfect cube After that, it was pretty easy to figure out that the smallest number that met these requirements was 2*2*2*3*3, or 72. That was a neat little exercise.
Interesting. I considered the prime factorization of n. If 2n is a perfect square, then 2 has an odd exponent in the factorization and all other exponents are even. Likewise, if 3n is a perfect cube, then 3's exponent is 2 mod 3 and all other exponents are multiples of three. So, 2's exponent is an odd multiple of 3. The smallest such number is 3. 3's exponent is even and 2 mod 3. The smallest such number is 2. All other exponents are even and multiples of three. The smallest possible value for them is zero. Therefore, n = 2³•3² = 72.
Given the first equation, we know n = 2^y*x^z, where y is odd and z is even. This is because 2^(y+1)*x^z = (2^((y+1)/2)*x^(z/2))^2, which is a perfect square (a perfect square times another perfect square will always be a perfect square). Given the second equation, we know n = 3^q*(r^w), where q is one less than a multiple of 3 and w is divisible by 3. This is because 3^(q+1)*r^w = (3^((q+1)/3)*r^(w/3))^3, which is a perfect cube (same logic as above but with perfect cubes). So we can then compare the n from the first equation to the n in the second equation: n = 2^y*x^z = 3^q*(r^w) At this point, it's obvious that if we want n to be as small as possible, we should make x=3 and r=2 n = 2^y*3^z = 3^q*2^w So y=w and z=q. The smallest positive integer that satisfies y being odd and w being divisible by 3 is 3. The smallest positive integer that satisfies z being even and q being one less than a multiple of 3 is 2. So n = 2^3*3^2 = 3^2*2^3 = 8*9 = 72. Checking our work: 2*72 = 144 = 12^2 3*72 = 216 = 6^3
wow a surprisingly simple one once you realize it needs 2 '3' factors and a '2' factor for every '3' plus one extra to fit the cube stuff. then since you have an even number of '3' factors and an odd number of '2' factors, that means when you add the next '2' factor by multiplying it by 2, each factor appears an even number of times, making it a square (144). so it is 3*3*2*2*2 = 72
Really fun problem. I solved it the same way as you did. After the video I also tried it starting from 3n: If 3n is a perfect cube, then n = 9k^3. Then we've got 3n = (3k)^3. We substitute it in 2n = 2(9k^3) = 18k^3 = q^2 The smallest k = 2. We get n = 9 * 2^3 = 9 * 8 = 72 so the same answer.
I'd use a different methods where I'd factor n into some list of prime factors (f1^a•f2^b•...). In order for 2•n to be perfect square, every prime factor has to appear an even number of times, except for 2 which appears once alone, so multiplying n by two gets a perfect square: 2•n=(2•f1^(2a)•f2^(2b)•...) Similarily, in order for 3n to be a perfect cube, n has to have the factor 3 appear at least twice: 3•n=3•(3•3•2•f1^a•f2^b•...) As we can see, in order for n to be a perfect cube n can not only have the factor 2 appear once, and must have it appear at least 3 times: 3•n=3•(3•3•2^3) is a perfect cube. Here we see that (2^3•3^2) still satisfies the first criteria, where 2•n=2•(2•2^2•3^2), and no number smaller than this can satisfy both criteria, so our smallest possible number must be n = 2^3•3^2 = 8•9 = 72
72? 5:51 We're cooking here... 6:55 Man I'm good, I did that in my head. Though granted, it was about my third answer, but it was the only one that stood up to the property in the premise.
Another solution: 2n is a perfect square and 3n is a perfect cube. 2n=y²; (2n)³=(y²)³; 8n³=y⁶ 3n=z³;(3n)²=(z³)²; 9n²=z⁶ So 8n³=y⁶ and 9n²=z⁶; 8n³ * 9n² = y⁶z⁶ 72n⁵=y⁶z⁶ 72n⁵=(yz)⁶; on the left side of this equation, it is immediately evident that n=72 because 72n⁵=72¹(72⁵)=72⁽¹⁺⁵⁾=72⁶, and 72⁶ satisfies the right side of the equation (yz)⁶ thus n=72 checking: 2n is a square because 2*72=144=12² 3n is a cube because 3*72=216=6³ QED
From the 2n=square and 3*n=cube conditions it follows (via Fundamental Theorem of Arithmetic) that n must have the form 2^(3a)*3^(3b)*q^(6c), a,b,q integer>0, c non-negative integer. The smallest such number is 2^3*3^2=72 (i.e. a=1,b=1, q=1).
n = 2^3.3^2, I have had this video in my feed a couple of times. I didn't have a clear solution yet this time it was obvious. Thank you for this elementary Number theory.
Hi! At 2:15 Canon get some intuition on how to get from “2n is perfect square” n = 2k^2. I mean - when I hear “2n is perfect square” I imagine 2n = k^2, so n = (k^2)/2
Hi, this confused me too but basically it works like this. 2n is a perfect square, so you can rewrite it as (√2n)^2 Perfect square means that it is an integer squared, so √2n must be an integer Rewrite as √2 * √n For that to be an integer, we need the √2 to be an integer and √n So you can substitute 2a for n and see what happens It becomes √2 * √2a which is 2* √a Almost an integer but it still has √a For √a to be an integer a would have to be a square of some number so the root would cancel So substitute with k^2 and it becomes √k^2 and then just k Putting that back together it becomes 2k, no more square roots So basically if you want 2n to be a perfect square then it’s root √2n must be an integer and for that to be an integer it must be some number 2k^2
*Here is a solution using exponents of primes...* 2^1 * n = 2^(1 + A) * 3^B 3^1 * n = 2^A * 3^(1 + B) n = 2^A * 3^B A must be threeven and odd. B must be even and congruent to 2 mod 3. Both must be at their minimum given the conditions above. This means A and B are 3 and 2 respectively. n = 2^3 * 3^2 = 72
Let's start with arguments...if 2n is a perfect square, then n must me even, and if 3n is a perfect cube, n must be a multiple of 3²...now wait if n is even and 3n is a perfect cube, n must be a multiple of 2³...in short n is a multiple of 2³×3²=72..still this is not general solution. Let's go back to the question, if n=72k, then 2n = 144k which is a perfect square, we conclude k is a perfect square. Again 3n=216k, which is a perfect cube, indicates that k is a perfect cube...bingo...solution of n is n=72k where k is both a perfect square and cube...now we may think of few such k...k= 1, 64, and so on... Great work and amazing video sir!!
A more longer way was to see that 2 | n and 6 | n so 12 | n. So n = 12 × k. Hence 24 × k must be a perfect square. Decomposing 24 you find the most straightforward decomposition which is 12 × 2 (could have also seen other ones and work with the pcm) and you deduce k = 6.
I had this approach: If 2n is a perfect square then n is divided by 2 so as to form the factor of 2^2. If 3n is a perfect cube then n is divided by 3^2=9 so as to form the factor of 3^3. Since n is divided both by 2 and by 9, n is divided by 18. Therefore n=18*k where k is a positive integer. 2n=2*18k=36*k=6^2*k and 3n=3*18k=54*k=3^3*2k If 2n is a perfect square then 6^2*k is a perfect square which means k is a perfect square. If 3n is a perfect cube then 3^3*2k is a perfect cube which means 2k is a perfect cube. Therefore k is divided by 2^2=4 so as to form the factor of 2^3. This means that k=4*z where z a positive integer. Meaning 2k=2*4z=2^3*z is a perfect cube so z must be a perfect cube.The smallest cube is 1. This concludes: z=1----> k=4*1 ----> k=4 ----> n=18*4 ----> n=72
So to "double number" get perfect square it has to have an odd number of 2's, to get a perfect cube by mult by 3, it needs to have 3n - 1 threes, and also 3m number of twos, so that means the smallest positive number that will work for is 2^3 * 3^2 or 8 * 9 = 72.
Nice video. I think you can follow this through to generalise the solution so that: n = 72t⁶ satisfies 2n=p² and 3n=q³ for all natural numbers t; n, p & q also being natural nunbers. You get this by following through that if 6k² =q³, k=6r and that r must be a perfect square and cube which is only possible if r is t⁶ where t is a natural number. So 3n = 216t⁶ or n =72t⁶. The first 3 n are 72 (the smallest n from the video), 4608, 52488.
This is the simplified version of a fifth grade problem given at the Regional Mathematics Olympiad in Romania, in 2024. Problem 3. Show that: a) there is an infinity of natural numbers n such that the number 2*n is a perfect square, and the number 3*n is a perfect cube; b) there is no natural number m such that the number 2+m is a perfect square and the number 3*m is a perfect cube.
I haven't watched the video yet, just wanted to give the solution that I found: Firstly, I decided to get rid of these disgusting things people call "words" and rewrited this problem as 2n=i²; 3n=k³. Then I started to experiment with those equations, took the corresponding roots from both sides and temporily omitted the "n"s. So, √2=i; ³√3=k. In theory, what I made was useless, but in practice, it helped me to think in the right way. So, then I looked to the ³√3 and realised that for there to be an integer, there must at least 2 more "3"s in the root, therefore the required number is divisible by 9. Then I looked at √2 and realised that n is also must be divisible by 2 (same thought processor, as in the last sentence), and because there was ³√3 in the second equation, then the n is surely divisible by 2³=8. After all that, it became bright that the n equals to the least common multiple of 8 and 9, i.e. 72.
Very nice problem ! If you're curious, you can also prove that the set of integers that satisfy these two properties (i.e. 2n=a^2 and 3n=b^3) is exactly equal to the set of integers of the form 72k^6, where k is a (strictly) positive integer ! And the minimum of that set is indeed 72 👌
Problem similar to yours with the 6 and 7 replacing your coefficients of 2 and 3: Find the smallest natural number n, such that 6n is a perfect square and 7n is a perfect cube. The smallest positive integer that satisfies the given conditions must be in the form n = 2^r∙3^s∙7^t r,s,t ∈N . The coefficients 6 and 7 of n determine this form. The smallest natural number is n = 2^3∙3^3∙7^2 = 10,584. 6n = 252^2 and 7n = 42^3
If you're going to brute force, brute forcing n directly is the wrong route. You want to brute force 3n. And while you're doing that, you want to make sure 3n is even. n has to be even, as all perfect squares are either odd, in which case 2n/2 is a fraction and thus 3n cannot be a perfect square, or even and of the form 4k, where k is a whole number. Since 3n needs to be even, you're basically looking for 6m, where m is n/2. Quickest way to brute force 3n (6m) is just to go through the perfect cubes until you find one that is divisible by 6 and check to see if 2/3 of that number is a perfect square. Obviously, 6³ is going to be the first to check, as no previous perfect cube is divisible by 6. 6³ = 6•6•6 = 216 216•2/3 = 72•2 = 144 √144 = 12 ✓ And we have a winner on our first pick. 2(72) = 12² and 3(72) = 6³, so n = 72. Another method, if you don't want to brute force it, is to predict the necessary prime factorization of n. Starting again with our resultant cube, n will need to be divisible by a multiple of 3 of the format 3^(3j-1). 3 will by necessity need to be a part of the prime factorization of 3n, as multiplying n by 3 adds 3 as a prime factor. As it needs to be a perfect cube, the number of factors of 3 in n needs to be one less than a number divisible by 3, so that when that 3 factor is added, we get 3n being divisible by 3^3j. Similarly, n also needs to be divisible by a multiple of 2 of the format 2^3k. As 3n does with 3, 2n will add a prime factor of 2, so n will need to already have 2 among its prime factors, and since 3n needs to be a perfect cube, the number of factors of 2 in n needs to be divisible by 3. Additionally, 3k+1 must be even, so that the number of factors of 2 in 2n is even (2^(3k+1)) while the number in 3n are still divisible by 3 (2^3k). Therefore k (and thus 3k) must be odd. The smallest number where the above conditions are met is where k = 1 and j = 1, with no additional prime factors: n = 2^3k•3^(3j-1) n = 2^(3(1))•3^(3(1)-1) n = 2³•3² n = 2•2•2•3•3 n = 8•9 = 72 2n = 144 = 2⁴•3² = (2²•3)² = 12² 3n = 216 = 2³•3³ = (2•3)³ = 6³
If 3n is a perfect cube, the minimum power of 2 in n has to be 3 and similarly minimum power of 3 in n will be 2 if 2n is a perfect square. Also, for the prime 2, on doing 2n, its power will become 3+1 = 4 which is a perfect square and similarly on doing 3n, the prime 3's power will become 2+1=3 which satisfies a cube's power...so the smallest number will be 2^3 × 3^2 = 72
Fun fact: 6,810,125,783,203,125,000,000,000,000,000 is the smallest number that "n" can be such that 2n is a perfect square, 3n is a perfect cube, and 5n is a perfect penteract (or 5th power).
Its clear that the number must be a composition of 2's and 3's, so it must be in the form 2^k * 3^w 2*2^k * 3^w is a perfect square, so k must be odd, and w even 3* 2^k * 3^w is a perfect cube, so k must be a multiple of 3. following the condition, the smallest pair of numbers is 3 and 2, 2^3 * 3^2 = 8*9 = 72
Prime factors have to be 2 and 3. Need one 2 so that 2n will be a square and two 3's so that 3n will be a cube. But the single 2 wouldn't work in the 3n situation, so we'll actually need three 2's so that 2n gives 2^4 x 3^2 and 3n gives 2^3 x 3^3. So 2^3 x 3^2 = 8 x 9 = 72.
N must be factorized as 2^(2k + 1) * 3^(3k - 1), with 3k - 1 being even for the square root of this number to be an integer, and 2k + 1 must be divisible by 3 so that the cube root of this number is an integer. If we substitute k = 1, then N = 2^3 * 3^2 = 72. Then 2N = 144 = 12^2, 3N = 216 = 6^3
My solution is so less elegant! 2n = k*k 3n = q*q*q 6n^2 = q(k*q)^2 Sqrt(6/q) * n = k*q Since 6/q must be an integer, q=6 is the solution. q^3= 216 so n =72
set 2n=a*a-(1),3n=b*b*b-(2),and we get 6n*n=a*a*b*b*b by multiplication, taking the square root, n(√6)=ab√b, since n,a,b are integers, we get n=ab and 6=b by comparing coefficients, plug in b=6 into (2) give us 3n=216,n=72.
actually, the fastest method is using the fact that 3n =q³ is a perfect cube, and since n is even, this perfect cube must also be even, and q must also be even. 2³ does not work, 4³ does not work, 6³ works. Done. And of course I also started with n=2*k², n= 3*3*l³ (q=3*l)=> 2|l and 3| k =>... buwhen you talked about the fasted method (compared to brute force) I got the idea of "inteligent force" not by going to test n=1,2,3.... but rather q=1,2,3,... which is much much faster
2n=k^2, 3n=m^3 -> n is divisible by 2, 3n=m^3 -> n is divisible by 2^3=8 and n is divisible by 3^3/3 = 9 -> n is divisible by 72. For n=72 we have 2*72=144=12^2 and 3*72=216=6^3.
If 2n = square, and 3n = cube, we know that n must be made up of 2s and 3s. Looking at the first line, the prime factorization of n must contain an odd number of 2s, and an even number of 3s. Looking at the second line, the number of 2s must be a multiple of 3, and the number of 3s must be a multiple of 3 -1. Therefore the smallest n must be 2*2*2*3*3. Though we have given the general solution to find as many as you like.
I don't. However, number theory only needs basic 8th grade knowledge. That's why it appeals to more people. If you've been on this channel long enough, you'd see that calculus is the predominant topic. Maybe it's time to do long lecture-style videos on algebra. 🤣
I don't speak english very well but in french the smallest integer positive for the question is zéro. But may be in english positive integer means positive and not zero.
It could also be solved in following way: as 2n is perfect square so n must have a factor of 2 so n=2k...(1) but as 3n is perfect cube so n must have a factor as 9 so n=9p ..(2) Now eliminate p and k with single variable n=2k=9p=18s...(3) •Combining two constraints 36s should be perfect a perfect square. 54s should be perfect cube Let s=4q(as 54s has incomplete triples of 2) •144q should be perfect a perfect square. 216q should be perfect cube. We could observe that the condition is fulfilled for any q=t⁶ 2n=144t⁶; n=72t⁶ is general solution t belongs to N.
From 2n = perfect square, n = must be an even number this is easy to get. From 3n = perfect cube, n must contain 3^2, and to be an even number it must contain a cube of an even number, smallest would be 2^3. hence 3^3 * 2^3 = 72.
2n is a perfect square n is s multiple of 8(=2³) to enable to have a perfect cube 3n is a perfect cube n is a multiple of 3² Therefore the smallest n is n=2³3²=72
2n=a^2 => n=2^(2k+1).x^2 (1) 3n=b^3 => n=3^(3k-1).y^3 (2) -because 2 and 3 are different numbers (obviously), we know n=2^(2k+1).3^(3k-1).u^6 -From (1) and (2) => 2^(2k+1)=p^3 =>2k+1=3m 3^(3k-1)=q^2 =>3k-1=2n (p,q are some factors of x,y) (k,m,n are intergers so that k,m,n>=0) =>k=2n-3m+2 => min(k)=1 when m=n=0 => n=2^3.3^2.u^6 =72.u^6 => min(n)=72 when u=1
Well, the absolute smallest n that meets those requirements is 0, but the thumbnail specifies it has to be positive. It also specifies that it has to be an integer, which is good, as it means I don't have to prove there aren't any non integers which satisfy this. All positive integers have a prime factorization. One way to express a prime factorization is as the product between all primes raised to some non-negative integer power. For instance, 1=2⁰3⁰5⁰7⁰11⁰... 12=2²3¹5⁰7⁰11⁰... 15=2⁰3¹5¹7⁰11⁰... A number is a perfect square if all its prime factor powers are divisible by 2. It's a perfect cube if all its prime factor powers are divisible by 3. If we say n=2^p2*3^p3*5^p5*... Then 2n=2^(p2+1)*3^p3*5^p5*... must be a square 3n=2^p2*3^(p3+1)*5^p5*... must be a cube So all powers except p2 must be even, all powers except p3 must be divisible by 3, p2+1 must be even, and p3+1 must be divisible by 3. p2 must be odd and divisible by 3 p3 must be even and be 2 mod 3 All other powers must be divisible by 6. Here, all powers can be computed independently. Therefore all powers should be the smallest non-negative integers that meet the above requirements. The smallest whole number divisible by 6 is 0, so all powers except p2 and p3 should be 0. p2 must be odd and divisible by 3. The smallest such number is 3. p3 must be even and 2 mod 3. The smallest such number is 2. n=2^3*3^2=8*9=72 n=72
Next viideo: a,b,n natural numbers, an is a perfect square and bn is a perfect cube. Does it hold in general that the solution will be: n=a(ab)^2. It hols for 3n a perfect square and 5n a perfect cube. :-)
The way I got it: 2n is a perfect square, so n must be a multiple of 2. 3n is a perfect cube, so n must be a multiple of 3. Therefore the smallest n has prime factors of 2 and 3 and nothing else. But to what powers? Well, 2n is a perfect square, so all its prime factors must be raised to even powers, so n must be 2 to an odd power times 3 to an even power. 3n is a perfect cube, so all its prime factors must be raised to a multiple of 3, so n must be 2 to a multiple of 3 times 3 to one less than a multiple of 3. The smallest odd multiple of 3 is 3. The smallest even number that's one less than a multiple of 3 is 2. Therefore n is 2 cubed times 3 squared, which is 8x9=72.
There are quite a lot of people who say they can't do maths. I think much of this is down to poor teaching. Your teaching is brilliant. The world needs more maths teachers like you. Never stop teaching!!!!!
Bang on! Maths was always my strongest subject but I saw so many that came to a point where they didn't understand something - usually something quite straightforward - and because it wasn't properly explained to them in language they could comprehend everything after that became a struggle and they effectively gave up. So sad because maths is such an elegant subject.
N is 72
If 2n is a perfect square, then N has an odd power of 2 as a factor.
If 3n is a perfect cube, then N has 2^(3A) as a factor and 3^(3B - 1) as a factor.
Putting A & B = 1, we get 2^3 * 3^2 = 72.
72*2=144=12^2
72*3=216=6^3
My same thought process once I saw the thumbnail.
My thought process exactly!! 😅😅
Yep. Took 20 seconds or so to figure out as a mental-math kata.
Your solution is very elegant.
I found the answer by thinking about prime factors, in a square the prime factors appear in pairs and in a cube they appear in triplets. The value must then have some number of prime factors as 2 and some as three. We can see that there must be an even amount of 3s and an odd amount of 2s. If we look at having only a single 2 as a prime factor and two 3s then it would be a perfect square when multiplied by 2 but not a perfect cube when multiplied by 3. If we instead have three 2s and two 3s as prime factors then it fulfil both conditions.
n = 2x2x2x3x3 = 72
2n = (2x2)(2x2)(3x3) = 144
3n = (2x2x2)(3x3x3) = 216
I found the solution using the prime factors..,put n=2^x*3^y.. then 2n=2^(x+1)*3^y.. therefore x is odd .. if we consider 3n, we have 3n=2^x*3^(y+1)..,therefore x must be multiple of 3..the smallest odd number which is multiple of 3 is 3.. then x=3.. furthermore y must be even because 2n is a square.. and y+1 must be a multiple of 3 because 3n is a cube.. the smallest number with these 2 ptoperties is 2..,therefore y=2.. finally we obtain n= 2^3*3^2=72
That's genius. We got reincarnation of Einstein before GTA6
nice find, this is how I did it as well
Infact all numbers 72*n where n is both a perfect square and cube satisfy this condition. 72, 4608, 294,912 etc
I used this method as well.
2n = a²
3n = b³
where a, b and n are positive integers. From here we get
6n = 3a²
6n = 2b³
This leads to the equation
3a² = 2b³ ⟹ a² = 2b³/3 ⟹ a = √(2b³/3).
Here obviously 2b³/3 is an integer meaning b has to be divisible by 3:
b = 3c ⟹ 2b³/3 = 2*3*3*c³
where c is positive integer. 2*3*3*c³ has to be a perfect square. This mean that the prime factorisation must have even copies of every prime factor. The smallest c to make that happen is
c = 2 ⟹ 2*3*3*c³ = 2*2*2*2*3*3 = 12² = a² = 2n ⟹ n = 12²/2 = 72.
I like this one!
@@ytsimontng Nice! Thanks!
Once found n=72, one can generate other numbers satisfying the conditions of 2n being perfect square and 3n being perfect cube by multiplying by p⁶, with p being a positive integer, since that factor is both a perfect square and a perfect cube.
Thus, the possible solutions are of the form n=72p⁶, with p=1 being of course the smallest. The next solution is n=4608 (with p=2) 💁🏻♂️
Excelent analysis. Simple and direct but most of all, logical. Thank and greetings from Perú
Great explanation. the solution you provide is simple and elegant.
Peace and good health to you, brother. Another fantastic video
You must realize that Fermat’s ‘last Theorem’ has a simpler proof within the scope of his level of math! Give it a go!
It made complete sense to me. A number is literally used as a given parameter here. In the same fashion you'd argue that the smallest pair (n,m) where 5*n = m^2 should be n=5=m if n and m are ints and greater then zero. Because it literally implies the 'other' n to also be 5 to become a perfect square.
Learning this number theory feels so fundamental. I am excited to explore it.
For 2n to be a perfect square n has to be even - divisible by 2. For 3n to be a perfect cube n has to be divisible by 9 and, since it’s even and 3 is not, by 8 as well. The smallest natural number that is divisible by both 8 and 9 is 8x9 = 72 - and it fits .
I don't understand that because it's a perfect cube it's divisible by 9 but because n is a perfect square the power of 3 must be even so 9
@@iskansinge7337 n is not a perfect square, 2n is
This is wrong on so many levels I have no idea where to begin
try 18
2x18 = 36 = 6²
3x18 = 64 = 4³
it must satisfy 2x2
it must satisfy 3x3x3
since 2 and 3 is already the coefficient of n,
so it only has to satisfy
2 and 3x3
so 18 it is
i understand what you're trying to say, but the problem can ONLY be solved with the smallest integer
Intuitively, you take 2 and 3 (the factors of n), you cube one and square the other, multiply one by the other: 8*9=72. It works because 2 and 3 are primes, and I'm pretty sure it works with other pairs of primes (not too sure about the powers involved, though, you probably need to jump into logs to prove it).
@@RegisMichelLeclerc nice!
Keep it up
A more elegant solution is to consider the factorization of n.
1) Since 2n is a square and 3n is a cube, the minimal n can only include factors of 2 and 3. So n=2^A*3^B.
2) Since 2n is a square, A must be odd and B even.
3) Since 3n is a cube, A must be a multiple of 3 and B a multiple of three minus one.
4) A is the smallest odd multiple of 3, that is 3 itself. B is the smallest even number that is one less than a multiple of 3, that is 2.
5) So n=2^3*3^2 = 72
You're doing the Lord's work.
Love your videos man, especially those dealing with numbers and number theory. Keep up this wonderful work of making math interesting and accessible. ❤❤❤
This is why I love number theory, it isn’t just mindless solving equations and with some thinking, the problem becomes easy, love the vid :)
A general solution would be:
if an = perfect square,
bn = pefect cube
solution:
n=ak^2
abk^2=q^3
k=ab
n=ak^2=a*a*a*b*b=a^3*b^2
The lowest integar value of n will always be n = a^3*b^2
Now, the big question would be... is there a value of n so that 3*(n^5) is a perfect cube and 5*(n^3) is a perfect power of 5, and if it's the case, how could you write and prove a generalisation?
At least if a and b are prime. If a=2^7 and b=3^7, then n=72 is still the smallest solution but it is smaller than a^3*b^2.
This is how I thought about it.
@@RegisMichelLeclerc I havent really gotten into proofs yet, I can only find a formula that seems correct
I solved by just looking at properties of n.
- Has only factors of 2 and 3, otherwise it would be unnecessarily bigger.
- Has a number of factors of 2 divisible by 3 (a factor of 8) and a number of factors of 3 divisible by 2 (a factor of 9).
- The count of factors of 2 is one less than a multiple of 2. The count of factors of 3 is one less than a multiple of 3. 8 and 9 satisfy this constraint already. Thus multiplying by 2 gives a perfect square, and multiplying by 3 gives a perfect cube.
Any number meeting these conditions is sufficient, and the smallest such number is 8 * 9 = 72.
I've been watching a fair number of your videos and I thoroughly enjoy your way of breaking things down. For someone whose interest in pure mathematics is far greater than current skill level, where would you suggest the person should start? Any introduction to pure mathematics you would recommend?
Math isn't boring, the way it's taught is and this is my proof
Thanks - I do run puzzle evenings in a local bar in England. The puzzles are mainly maths based. This is a great one!
Once you get to 6k^2 =q^3 you can strengthen your immediate jump to k=6 by pointing out that 6 is square-free and so there are no squares or cubes lurking in the factorization of 6.
Elegantly done ...again! I'm not going to admit to trying to 'count' my way to solution.
😂
2n is a perfect square => n is even
3n is a perfect cube mean => a) n contains at least the cube of 2 which is 8
b) n contains at least the square of 3 which is 9
Smallest possible value of n is 8*9=72
Good job 👍
Permit me to solve it differently.
There is a very intuitive method which requires no algebra:
Consider the characteristics of perfect squares and cubes. A perfect square has only prime factors that come in pairs (such as 2x2) while a perfect cubes has only prime factors which come in triplets (2x2x2). When we add a single 2 to the prime factorization of "n", we get a perfect square, so "n" must already have at least one 2 for that new 2 to pair with. Likewise, n must have at least two 3s. Okay, we now know that "n" has 2s and 3s in its factorization, and if we add a single 2, all the factors will be in pairs, and if we add a single 3, all the factors will be in triplets. Effectively that means that the number of 2s is a multiple of 3 which is one less that a multiple of 2, while the number of 3s is a multiple of 2 which is one less then a multiple of 3. Three 2s and two 3s meets these conditions, and if we starts removing any pairs or triplets of factors then we'll either run out of 2s or 3s, and we've already established that "n" has 2s and 3s in it, so this is also the smallest factorization which will work, and 2x2x2x3x3 is 72.
And honestly, this is all belabouring the point quite a bit. Once you start thinking about the prime factorizations and how squares and cubes work, the answer just kind of falls out.
Consider the prime factorization of n. If n has a factor of any prime other than 2 or 3, then we could make a smaller number by taking out all factors of that prime. So, the only prime factors are 2 and 3. So, we define a and b such that n = (2^a) (3^b).
Note that 2n has to be a perfect square. So 2^(a+1) 3^b is a perfect square, and hence a+1 and b are both even. If a+1 is even, then a is odd.
Furthermore, note that 3n has to be a perfect cube. So 2^a 3^(b+1) is a perfect cube and hence a and b+1 are both multiples of 3. If b+1 is a multiple of 3, then b = 3k+2 for some integer k.
The smallest odd multiple of 3 is 3, so a is 3. The smallest even number of the form 3k+2 is 2, so b is 2. Thus, the smallest value of n is (2^3) (3^2) = (8)(9) = 72
I knew that initial insight, then I just grew the sequences starting with 3 x 2 for the 2n is square, and 2 x 3 for the 3n is cubed. Grown in tandem. Easy.
@@Necrozene I wanted to be thorough, but also make it so that the process was followable by his audience. I think I succeeded?
@@chaosredefined3834 Yes, very nice as always. My method was very sloppy, and to explain how and why it would work to someone else would have got me into tangles.
The ideology you used is also how you prove Root "N" Is a irrational number, Perfect and simple application of divisibility of integers.
to be honest i wasnt able to understand the first notion of 4k² = 2n , BUT what i understood based on the statement was 2 times n is a perfect square, which means if the perfect square itself was being square root it should be equal to 2 times n , 2.n = square root of perfect square variable , that being said the perfect square variable is equal to (2.n)^2 , and we can substitute the variable with letter x = 4.n^2
If 2n=i^2 and 3n=j^3, we can deduce that ‘i’ can only be an even value due to ‘n’ being restricted to integers and being multiplied by an even integer. We can also deduce that ‘j’ must be a multiple of 3 for the same reasoning. From the given equations, we also know i^2/2=j^3/3=n. After this, we could check every multiple of 6 to replace ‘i’ or ‘j’ and solve for the other, and I would recommend testing inputs for ‘j’ since it has the potential to grow larger results faster. It just so happens j=6 results in a i=12 & n=72, which since ‘i’ and ‘j’ are integers, they create their respective square and cube.
Oh, my memories of a white chalk and a (green) blackboard!
You are a great teacher
Never Stop Teaching!
I did it a slightly different way by figuring out that the prime factorization of n would require:
- an odd number of 2's and an even number of 3's, in order for 2n to be a perfect square
- a number of 3's congruent to 2 mod 3, and a number of 2's that was a multiple of 3, in order for 3n to be a perfect cube
After that, it was pretty easy to figure out that the smallest number that met these requirements was 2*2*2*3*3, or 72. That was a neat little exercise.
Interesting. I considered the prime factorization of n. If 2n is a perfect square, then 2 has an odd exponent in the factorization and all other exponents are even. Likewise, if 3n is a perfect cube, then 3's exponent is 2 mod 3 and all other exponents are multiples of three.
So, 2's exponent is an odd multiple of 3. The smallest such number is 3.
3's exponent is even and 2 mod 3. The smallest such number is 2.
All other exponents are even and multiples of three. The smallest possible value for them is zero.
Therefore, n = 2³•3² = 72.
Given the first equation, we know n = 2^y*x^z, where y is odd and z is even.
This is because 2^(y+1)*x^z = (2^((y+1)/2)*x^(z/2))^2, which is a perfect square (a perfect square times another perfect square will always be a perfect square).
Given the second equation, we know n = 3^q*(r^w), where q is one less than a multiple of 3 and w is divisible by 3.
This is because 3^(q+1)*r^w = (3^((q+1)/3)*r^(w/3))^3, which is a perfect cube (same logic as above but with perfect cubes).
So we can then compare the n from the first equation to the n in the second equation:
n = 2^y*x^z = 3^q*(r^w)
At this point, it's obvious that if we want n to be as small as possible, we should make x=3 and r=2
n = 2^y*3^z = 3^q*2^w
So y=w and z=q.
The smallest positive integer that satisfies y being odd and w being divisible by 3 is 3.
The smallest positive integer that satisfies z being even and q being one less than a multiple of 3 is 2.
So n = 2^3*3^2 = 3^2*2^3 = 8*9 = 72.
Checking our work:
2*72 = 144 = 12^2
3*72 = 216 = 6^3
If we replace 2 and 3 from the initial question with A and B respectively, where A and B are coprime integers, then the solution is always A³ * B². (:
wow a surprisingly simple one once you realize it needs 2 '3' factors and a '2' factor for every '3' plus one extra to fit the cube stuff. then since you have an even number of '3' factors and an odd number of '2' factors, that means when you add the next '2' factor by multiplying it by 2, each factor appears an even number of times, making it a square (144). so it is 3*3*2*2*2 = 72
Really fun problem. I solved it the same way as you did.
After the video I also tried it starting from 3n:
If 3n is a perfect cube, then n = 9k^3. Then we've got 3n = (3k)^3.
We substitute it in 2n = 2(9k^3) = 18k^3 = q^2
The smallest k = 2. We get n = 9 * 2^3 = 9 * 8 = 72 so the same answer.
I'd use a different methods where I'd factor n into some list of prime factors (f1^a•f2^b•...).
In order for 2•n to be perfect square, every prime factor has to appear an even number of times, except for 2 which appears once alone, so multiplying n by two gets a perfect square:
2•n=(2•f1^(2a)•f2^(2b)•...)
Similarily, in order for 3n to be a perfect cube, n has to have the factor 3 appear at least twice:
3•n=3•(3•3•2•f1^a•f2^b•...)
As we can see, in order for n to be a perfect cube n can not only have the factor 2 appear once, and must have it appear at least 3 times:
3•n=3•(3•3•2^3) is a perfect cube.
Here we see that (2^3•3^2) still satisfies the first criteria, where
2•n=2•(2•2^2•3^2), and no number smaller than this can satisfy both criteria, so our smallest possible number must be n = 2^3•3^2 = 8•9 = 72
I like your explanation.
2n is a square
So take n=2p^2
3n=3×2p^2=6p^2=6×6×6=6^3
p=6,n=2×6^2=72
3n=3×72=216=6^3.
2n=144÷12^2.
72?
5:51 We're cooking here...
6:55 Man I'm good, I did that in my head. Though granted, it was about my third answer, but it was the only one that stood up to the property in the premise.
sir ur best
Another solution:
2n is a perfect square and 3n is a perfect cube.
2n=y²; (2n)³=(y²)³; 8n³=y⁶
3n=z³;(3n)²=(z³)²; 9n²=z⁶
So 8n³=y⁶ and 9n²=z⁶;
8n³ * 9n² = y⁶z⁶
72n⁵=y⁶z⁶
72n⁵=(yz)⁶; on the left side of this equation, it is immediately evident that n=72 because 72n⁵=72¹(72⁵)=72⁽¹⁺⁵⁾=72⁶, and 72⁶ satisfies the right side of the equation (yz)⁶
thus n=72
checking: 2n is a square because 2*72=144=12²
3n is a cube because 3*72=216=6³
QED
From the 2n=square and 3*n=cube conditions it follows (via Fundamental Theorem of Arithmetic) that n must have the form 2^(3a)*3^(3b)*q^(6c), a,b,q integer>0, c non-negative integer. The smallest such number is 2^3*3^2=72 (i.e. a=1,b=1, q=1).
Extremely clean!
n = 2^3.3^2, I have had this video in my feed a couple of times. I didn't have a clear solution
yet this time it was obvious.
Thank you for this elementary Number theory.
Hi! At 2:15 Canon get some intuition on how to get from “2n is perfect square” n = 2k^2. I mean - when I hear “2n is perfect square” I imagine 2n = k^2, so n = (k^2)/2
Hi, this confused me too but basically it works like this.
2n is a perfect square, so you can rewrite it as (√2n)^2
Perfect square means that it is an integer squared, so √2n must be an integer
Rewrite as √2 * √n
For that to be an integer, we need the √2 to be an integer and √n
So you can substitute 2a for n and see what happens
It becomes √2 * √2a which is 2* √a
Almost an integer but it still has √a
For √a to be an integer a would have to be a square of some number so the root would cancel
So substitute with k^2 and it becomes √k^2 and then just k
Putting that back together it becomes 2k, no more square roots
So basically if you want 2n to be a perfect square then it’s root √2n must be an integer and for that to be an integer it must be some number 2k^2
@@qloxer1264 NowIAmConfusedMuchMore.jpg :)
What about finding all natural ns meeting these conditions?
Taking this a step further, are there larger numbers that would satisfy the two equations?
*Here is a solution using exponents of primes...*
2^1 * n = 2^(1 + A) * 3^B
3^1 * n = 2^A * 3^(1 + B)
n = 2^A * 3^B
A must be threeven and odd.
B must be even and congruent to 2 mod 3.
Both must be at their minimum given the conditions above.
This means A and B are 3 and 2 respectively.
n = 2^3 * 3^2 = 72
נחמד מאד, לא הכרתי. תודה!
Let's start with arguments...if 2n is a perfect square, then n must me even, and if 3n is a perfect cube, n must be a multiple of 3²...now wait if n is even and 3n is a perfect cube, n must be a multiple of 2³...in short n is a multiple of 2³×3²=72..still this is not general solution. Let's go back to the question, if n=72k, then 2n = 144k which is a perfect square, we conclude k is a perfect square. Again 3n=216k, which is a perfect cube, indicates that k is a perfect cube...bingo...solution of n is n=72k where k is both a perfect square and cube...now we may think of few such k...k= 1, 64, and so on...
Great work and amazing video sir!!
Quick way is to count up in sixes and check out the squares you come to - i.e., find the candidate for 2n.
A more longer way was to see that 2 | n and 6 | n so 12 | n. So n = 12 × k. Hence 24 × k must be a perfect square. Decomposing 24 you find the most straightforward decomposition which is 12 × 2 (could have also seen other ones and work with the pcm) and you deduce k = 6.
I had this approach:
If 2n is a perfect square then n is divided by 2 so as to form the factor of 2^2. If 3n is a perfect cube then n is divided by 3^2=9 so as to form the factor of 3^3. Since n is divided both by 2 and by 9, n is divided by 18. Therefore n=18*k where k is a positive integer.
2n=2*18k=36*k=6^2*k and 3n=3*18k=54*k=3^3*2k
If 2n is a perfect square then 6^2*k is a perfect square which means k is a perfect square. If 3n is a perfect cube then 3^3*2k is a perfect cube which means 2k is a perfect cube. Therefore k is divided by 2^2=4 so as to form the factor of 2^3.
This means that k=4*z where z a positive integer. Meaning 2k=2*4z=2^3*z is a perfect cube so z must be a perfect cube.The smallest cube is 1.
This concludes:
z=1----> k=4*1 ----> k=4 ----> n=18*4 ----> n=72
Who else just did this through trial and error by comparing both perfect squares and perfect cubes?
Wow, amazing!
So to "double number" get perfect square it has to have an odd number of 2's, to get a perfect cube by mult by 3, it needs to have 3n - 1 threes, and also 3m number of twos, so that means the smallest positive number that will work for is 2^3 * 3^2 or 8 * 9 = 72.
Nice video. I think you can follow this through to generalise the solution so that: n = 72t⁶ satisfies 2n=p² and 3n=q³ for all natural numbers t; n, p & q also being natural nunbers. You get this by following through that if 6k² =q³, k=6r and that r must be a perfect square and cube which is only possible if r is t⁶ where t is a natural number. So 3n = 216t⁶ or n =72t⁶. The first 3 n are 72 (the smallest n from the video), 4608, 52488.
This is the simplified version of a fifth grade problem given at the Regional Mathematics Olympiad in Romania, in 2024.
Problem 3. Show that:
a) there is an infinity of natural numbers n such that the number 2*n is a perfect square, and the number 3*n is a perfect cube;
b) there is no natural number m such that the number 2+m is a perfect square and the number 3*m is a perfect cube.
I haven't watched the video yet, just wanted to give the solution that I found:
Firstly, I decided to get rid of these disgusting things people call "words" and rewrited this problem as
2n=i²;
3n=k³.
Then I started to experiment with those equations, took the corresponding roots from both sides and temporily omitted the "n"s. So,
√2=i;
³√3=k.
In theory, what I made was useless, but in practice, it helped me to think in the right way. So, then I looked to the ³√3 and realised that for there to be an integer, there must at least 2 more "3"s in the root, therefore the required number is divisible by 9. Then I looked at √2 and realised that n is also must be divisible by 2 (same thought processor, as in the last sentence), and because there was ³√3 in the second equation, then the n is surely divisible by 2³=8. After all that, it became bright that the n equals to the least common multiple of 8 and 9, i.e. 72.
Very nice problem !
If you're curious, you can also prove that the set of integers that satisfy these two properties (i.e. 2n=a^2 and 3n=b^3) is exactly equal to the set of integers of the form 72k^6, where k is a (strictly) positive integer ! And the minimum of that set is indeed 72 👌
Problem similar to yours with the 6 and 7 replacing your coefficients of 2 and 3:
Find the smallest natural number n, such that 6n is a perfect square and 7n is a perfect cube.
The smallest positive integer that satisfies the given conditions must be in the form
n = 2^r∙3^s∙7^t r,s,t ∈N . The coefficients 6 and 7 of n determine this form.
The smallest natural number is
n = 2^3∙3^3∙7^2 = 10,584.
6n = 252^2 and 7n = 42^3
If you're going to brute force, brute forcing n directly is the wrong route. You want to brute force 3n. And while you're doing that, you want to make sure 3n is even. n has to be even, as all perfect squares are either odd, in which case 2n/2 is a fraction and thus 3n cannot be a perfect square, or even and of the form 4k, where k is a whole number. Since 3n needs to be even, you're basically looking for 6m, where m is n/2.
Quickest way to brute force 3n (6m) is just to go through the perfect cubes until you find one that is divisible by 6 and check to see if 2/3 of that number is a perfect square. Obviously, 6³ is going to be the first to check, as no previous perfect cube is divisible by 6.
6³ = 6•6•6 = 216
216•2/3 = 72•2 = 144
√144 = 12 ✓
And we have a winner on our first pick. 2(72) = 12² and 3(72) = 6³, so n = 72.
Another method, if you don't want to brute force it, is to predict the necessary prime factorization of n. Starting again with our resultant cube, n will need to be divisible by a multiple of 3 of the format 3^(3j-1). 3 will by necessity need to be a part of the prime factorization of 3n, as multiplying n by 3 adds 3 as a prime factor. As it needs to be a perfect cube, the number of factors of 3 in n needs to be one less than a number divisible by 3, so that when that 3 factor is added, we get 3n being divisible by 3^3j.
Similarly, n also needs to be divisible by a multiple of 2 of the format 2^3k. As 3n does with 3, 2n will add a prime factor of 2, so n will need to already have 2 among its prime factors, and since 3n needs to be a perfect cube, the number of factors of 2 in n needs to be divisible by 3. Additionally, 3k+1 must be even, so that the number of factors of 2 in 2n is even (2^(3k+1)) while the number in 3n are still divisible by 3 (2^3k). Therefore k (and thus 3k) must be odd.
The smallest number where the above conditions are met is where k = 1 and j = 1, with no additional prime factors:
n = 2^3k•3^(3j-1)
n = 2^(3(1))•3^(3(1)-1)
n = 2³•3²
n = 2•2•2•3•3
n = 8•9 = 72
2n = 144 = 2⁴•3² = (2²•3)² = 12²
3n = 216 = 2³•3³ = (2•3)³ = 6³
Cubes grow faster than squares therefore n is small, simple search upwards yields the right answer in no time.
If 3n is a perfect cube, the minimum power of 2 in n has to be 3 and similarly minimum power of 3 in n will be 2 if 2n is a perfect square. Also, for the prime 2, on doing 2n, its power will become 3+1 = 4 which is a perfect square and similarly on doing 3n, the prime 3's power will become 2+1=3 which satisfies a cube's power...so the smallest number will be 2^3 × 3^2 = 72
Fun fact: 6,810,125,783,203,125,000,000,000,000,000 is the smallest number that "n" can be such that 2n is a perfect square, 3n is a perfect cube, and 5n is a perfect penteract (or 5th power).
n = 72k^6 is the general solution for all solutions, where k is a positive integer, the smallest one being 72 where k=1
Its clear that the number must be a composition of 2's and 3's, so it must be in the form 2^k * 3^w
2*2^k * 3^w is a perfect square, so k must be odd, and w even
3* 2^k * 3^w is a perfect cube, so k must be a multiple of 3.
following the condition, the smallest pair of numbers is 3 and 2, 2^3 * 3^2 = 8*9 = 72
Prime factors have to be 2 and 3. Need one 2 so that 2n will be a square and two 3's so that 3n will be a cube. But the single 2 wouldn't work in the 3n situation, so we'll actually need three 2's so that 2n gives 2^4 x 3^2 and 3n gives 2^3 x 3^3. So 2^3 x 3^2 = 8 x 9 = 72.
72 / 2*72=144 srt=12 / 3*72=216 crt=6 // next value is 4608
N must be factorized as 2^(2k + 1) * 3^(3k - 1), with 3k - 1 being even for the square root of this number to be an integer, and 2k + 1 must be divisible by 3 so that the cube root of this number is an integer. If we substitute k = 1, then N = 2^3 * 3^2 = 72. Then 2N = 144 = 12^2, 3N = 216 = 6^3
2 3s and 3 2s gives an even number and 3 of each factor when applied
My solution is so less elegant!
2n = k*k
3n = q*q*q
6n^2 = q(k*q)^2
Sqrt(6/q) * n = k*q
Since 6/q must be an integer, q=6 is the solution. q^3= 216 so n =72
A person teaching Maths WITH GOOD HANDWRITING?????
set 2n=a*a-(1),3n=b*b*b-(2),and we get 6n*n=a*a*b*b*b by multiplication, taking the square root, n(√6)=ab√b, since n,a,b are integers, we get n=ab and 6=b by comparing coefficients, plug in b=6 into (2) give us 3n=216,n=72.
actually, the fastest method is using the fact that 3n =q³ is a perfect cube, and since n is even, this perfect cube must also be even, and q must also be even. 2³ does not work, 4³ does not work, 6³ works. Done.
And of course I also started with n=2*k², n= 3*3*l³ (q=3*l)=> 2|l and 3| k =>... buwhen you talked about the fasted method (compared to brute force) I got the idea of "inteligent force" not by going to test n=1,2,3.... but rather q=1,2,3,... which is much much faster
i have to watch this again
2n=k^2, 3n=m^3 -> n is divisible by 2, 3n=m^3 -> n is divisible by 2^3=8 and n is divisible by 3^3/3 = 9 -> n is divisible by 72. For n=72 we have 2*72=144=12^2 and 3*72=216=6^3.
If 2n = square, and 3n = cube, we know that n must be made up of 2s and 3s. Looking at the first line, the prime factorization of n must contain an odd number of 2s, and an even number of 3s.
Looking at the second line, the number of 2s must be a multiple of 3, and the number of 3s must be a multiple of 3 -1.
Therefore the smallest n must be 2*2*2*3*3. Though we have given the general solution to find as many as you like.
Why do you like the number theory more than any other domain in mathematics ?
I don't. However, number theory only needs basic 8th grade knowledge. That's why it appeals to more people. If you've been on this channel long enough, you'd see that calculus is the predominant topic. Maybe it's time to do long lecture-style videos on algebra. 🤣
I don't speak english very well but in french the smallest integer positive for the question is zéro. But may be in english positive integer means positive and not zero.
It could also be solved in following way:
as 2n is perfect square so n must have a factor of 2 so n=2k...(1) but as 3n is perfect cube so n must have a factor as 9 so n=9p ..(2)
Now eliminate p and k with single variable n=2k=9p=18s...(3)
•Combining two constraints
36s should be perfect a perfect square.
54s should be perfect cube
Let s=4q(as 54s has incomplete triples of 2)
•144q should be perfect a perfect square.
216q should be perfect cube.
We could observe that the condition is fulfilled for any q=t⁶
2n=144t⁶;
n=72t⁶ is general solution t belongs to N.
👌👍
How about the 18th century proof for Fermat’s Last?
From 2n = perfect square, n = must be an even number this is easy to get. From 3n = perfect cube, n must contain 3^2, and to be an even number it must contain a cube of an even number, smallest would be 2^3. hence 3^3 * 2^3 = 72.
As 3n is a perfect cube, n should contain 3^2 and 2^3
As 2n is perfect square, n should contain 3^2and 2^3
Hence n= 3^2*2^3=72
n=0
i was wondering why this was so complicated, then i realized 2n = perf sq. and 3n = perf. cube was talking about the same n to satisfy both.
I just checked all of the perfect cubes between 1 and 216. It turns out that you didn't need to specify that n is a natural number.
2n is a perfect square
n is s multiple of 8(=2³) to enable to have a perfect cube
3n is a perfect cube
n is a multiple of 3²
Therefore the smallest n is n=2³3²=72
2n=a^2 => n=2^(2k+1).x^2 (1)
3n=b^3 => n=3^(3k-1).y^3 (2)
-because 2 and 3 are different numbers (obviously), we know n=2^(2k+1).3^(3k-1).u^6
-From (1) and (2)
=> 2^(2k+1)=p^3 =>2k+1=3m
3^(3k-1)=q^2 =>3k-1=2n
(p,q are some factors of x,y)
(k,m,n are intergers so that k,m,n>=0)
=>k=2n-3m+2
=> min(k)=1 when m=n=0
=> n=2^3.3^2.u^6
=72.u^6
=> min(n)=72 when u=1
Well, the absolute smallest n that meets those requirements is 0, but the thumbnail specifies it has to be positive. It also specifies that it has to be an integer, which is good, as it means I don't have to prove there aren't any non integers which satisfy this.
All positive integers have a prime factorization. One way to express a prime factorization is as the product between all primes raised to some non-negative integer power. For instance, 1=2⁰3⁰5⁰7⁰11⁰...
12=2²3¹5⁰7⁰11⁰...
15=2⁰3¹5¹7⁰11⁰...
A number is a perfect square if all its prime factor powers are divisible by 2. It's a perfect cube if all its prime factor powers are divisible by 3.
If we say n=2^p2*3^p3*5^p5*...
Then
2n=2^(p2+1)*3^p3*5^p5*... must be a square
3n=2^p2*3^(p3+1)*5^p5*... must be a cube
So all powers except p2 must be even, all powers except p3 must be divisible by 3, p2+1 must be even, and p3+1 must be divisible by 3.
p2 must be odd and divisible by 3
p3 must be even and be 2 mod 3
All other powers must be divisible by 6.
Here, all powers can be computed independently. Therefore all powers should be the smallest non-negative integers that meet the above requirements.
The smallest whole number divisible by 6 is 0, so all powers except p2 and p3 should be 0.
p2 must be odd and divisible by 3. The smallest such number is 3.
p3 must be even and 2 mod 3. The smallest such number is 2.
n=2^3*3^2=8*9=72
n=72
I'm not sure I caught that properly. if 2n is a perfect square, shouldn't that imply that n=½k² instead of 2k²?
Im really impressed
Next viideo: a,b,n natural numbers, an is a perfect square and bn is a perfect cube. Does it hold in general that the solution will be: n=a(ab)^2. It hols for 3n a perfect square and 5n a perfect cube. :-)
n=2^3•3^2
2n=2^4•3^2=12^2
3n=2^3•3^3=6^3
The way I got it:
2n is a perfect square, so n must be a multiple of 2.
3n is a perfect cube, so n must be a multiple of 3.
Therefore the smallest n has prime factors of 2 and 3 and nothing else. But to what powers?
Well, 2n is a perfect square, so all its prime factors must be raised to even powers, so n must be 2 to an odd power times 3 to an even power.
3n is a perfect cube, so all its prime factors must be raised to a multiple of 3, so n must be 2 to a multiple of 3 times 3 to one less than a multiple of 3.
The smallest odd multiple of 3 is 3. The smallest even number that's one less than a multiple of 3 is 2. Therefore n is 2 cubed times 3 squared, which is 8x9=72.