Hi, For fun: 1 "the next thing that we want to do", 1 "what I want to notice", 1 "so now what we are going to do", 2 "I'll go ahead and", 2 "so let's go ahead and", 1 "so let's may be go ahead and", 1 "I can go ahead and", 1 "we can go ahead and", 1 "and so on and so forth", 1 "great", 6:05 : a good place to look at. 6:31 : 2b or not 2b, that is the question. 19:48 : whouah! whouah !
A trick I found to make the solution quicker is that since n² has the same parity as a²b², we can use (ab - 2) as an upper bound for n, which leads to a more restrictive equation than with (ab - 1) and leaves us with only a handful of cases to check.
@@PlayerMathinson 4(a+b) is alway even. If a²b² is even the same is a²b²-4(a+b)=n². If a²b² is odd a²b²-4(a+b)=n² is odd. Thus they have the same parity
Apparently my first comment did NOT appear for other people so let’s try again... 5:01 Good inequality to stop 6:01 Good place to look at 20:07 Sin áit mhaith le stopadh Daily... How many elements do the following sets have : A = {x ∈ ℚ | x = (n² + 1)/(2n² + n + 1), n ∈ ℕ, n ∈ [1, 1000]} B = {x ∈ ℚ | x = (n² + 1)/(2n² + n + 1), n ∈ ℕ} Posting hint in another comment.
Hint... To be clear, we’re talking about sets, not multisets. So we look for distincts elements in the sets and you should look in what cases duplicates happen.
As none of the elements of A or B are zero we can consider their reciprocals minus 2. A reciprocal of a fraction as above can be written as 2+(n-1)/(n^2+1), so we have that the reciprocals minus 2 have the form (n-1)/(n^2+1), for appropriate values of n, or as m/(m^2+2m+2) for m=n-1. Taking another reciprocal (if m is not zero, thus n is not 1) and subtracting 2 yields m+2/m, whereas when m=0 the m/(m^2+2m+2)=0. So we need to see how many nonzero values m+2/m can assume when m lies in [0,999]. The function x+2/x is monotonically increasing for x>sqrt(2), so for m>=2 m+2/m will not be repeating values, and will increase for increasing m. When m=1 m+2/m=3, when m=2 it's also 3, so there is exactly one repeated value. Moreover 0 is never reached. Thus the cardinality of A is 1000-1=999. The same argument shows that the cardinality of B is countably infinite. EDIT: I also need to remark that the operation of taking the reciprocal and subtracting 2 is injective. This is hopefully obvious/well known. EDIT2: There is a very simple way to see that the cardinality of B is infinite while doing barely no work: Notice how the function (x^2+1)/(2x^2+x+1) is rational, and thus algebraic. We'll show that *No nonconstant algebraic functions can assume one value infinitely many times.* As a bonus we'll get that sin(x) is not algebraic :). The proof is extremely easy: say f(x) is algebraic, and p(x,f(x))=0 for some polynomial p. If f(x)=y for a fixed y then p(x,y)=0. As an equation of x, that is taking y as a fixed parameter, we can see that if it has infinitely many solutions then the polynomial must be constant in x. But then y must be constant. This concludes the proof of the mini-fact we'll use. Since no values can be attained infinitely many times the image of an infinite set must be infinite (of the same cardinality). This concludes the solution.
For common solution we must have n² + 1 and (n² + 1)+n² + n have a common factor. Hence n² + 1 and n² + n must have common factor which happens only fo n = 1. Hence 999 elements.
A rock star in math because even only 5% of math PROFESSORS(!) truly succeed in mathematical competitions like this. Studying maths in university is up to 1.5 orders of magnitude easier than this. Yes, you need to know much more than here but you need far less mathematical fantasy.
@@howmathematicianscreatemat9226 yeah, we don't have to use our brain this much to solve questions like these which require creativity & out of the box thinking sometimes. Sometimes understanding, remembering & regurgitating those truckloads of abstract theorems itself is a difficult task at university, let alone APPLYING them in creative ways to solve challenging problems 😅
They are really freaking cool honestly, and there is a certain satisfaction in doubling back and making a full loop check even when you are sure of how you arrived at the answer. Finding that n is the same both in the algebra of the subcases and by substituting a and b from the subcase by which you arrived at an n into the original expresssion and coming up with that same n again is fundamentally sublime. You just kick back, and enjoy a nirvana of "Man, math just works" Not that I'm exceptional at it, but it goes w/o saying- math logic is remarkably better than any other because it's so fundamental.
There's a small shortcut for the a=2 case. Note that (2b-1)^2=n^2+3^2 is a Pythagorean triple with one side length of 3, implying n=4 and 2b-1=5. Although is does make it harder to realize the 0+9=9 case is actually relevant.
My thoughts how to solve this: since (ab)^2 is already a square, we need to look at the difference between that and the perfect square just below it. Once a and b get beyond a certain since, this difference is going to be > 4(a+b) and therefore you only really need to test small values of a and b for that to hold. When I solved this , I consider 0 NOT a Natural number. Natural numbers should be 1, 2, 3, .....
case 1 has also (trivial) subcase 3: ((b-2)-n)((b-2)+n) = 8 may have both outer parenthesis to be negative. But that is possible only for b < 2, thus we should quickly check (and rule out) a=1, b=1.
Fun problem! But the parity argument (as mentioned below) plus some factoring significantly simplifies the cases to check. I also explored a=b in general which leads to the interesting fact that a^4-8a = a*(a-2)*(a^2+2a+4) is square ... but isn't much help in this problem but might be a useful approach for similar problems. (edited)
@@minime1235able That was a stupid error. Ugh! Thanks for the correction. My only excuse is that I approached the cases very differently and didn't put in the diligence when I followed his method. I started with the parity argument to realize that n= n^2 you derive ab (b-2)^2 = n^2 +8 But 1 and 9 are the only perfect squares that differ by exactly eight [Recall that (k+2m)^2 - k^2 = 4km + 4m^2 = 4(k+m)m but this is only 8 if k=m=1] Which yields (a,b)=(1,5) and its symmetric pair.
1. if a = 1 we have b^2-4*b-4 it has more than 1 root, thus can't be square but can be equal to 1 when x = 5. [1,5],[5,1]. 2 if a>=b>1 then a*b > a+ b or a+b = a*b-k with k>=1, we're getting x^2 - 4*x + 4*k which is always > 1 but can have one root when k = 1. so a*b = a+b-1 which can only be true in [2,3],[3,2]
case a=1: If (b-2)^2 = n^2+8 then you have 2 perfect squares which differ by 8. For (16,25) and above there are no solutions. So you look at all couples below that. And it's quickly seen that only (1,9) fits the bill. Similarly for a=2, but this time since the difference is 9 you include(16,25) to be checked which gives a solution.
@@leif1075 Partway through manipulating b^2-4(b+1)=n^2 he gets (b-2)^2-n^2=8. But the difference of successive squares (to say nothing of more widely separated squares) grows quickly enough to rapidly eliminate the possibility of such a difference being as low as 8. At 5^2-4^2 it's already larger. Checking squares lower than 5^2 (i.e. 1, 4, 9, and 16) against each other, the only pair that differs by 8 is (1,9). So 9=(b-2)^2 in that solution.
factorize b^2·(a + 2·(√(1 - b^3) + 1)/b^2)·(a - 2·(√(1 - b^3) - 1)/b^2) = n^2 u can see for all values greater than b =1 , we are going to have complex number. if b= 1 then (a+2)^2 = n^2 or a = n -2; if b = 1 then b = n-2; or as u said, a*b
Wow! Nice problem! I can’t really say that it was hard, but I have to admit that I enjoyed watching you explain it. Keep up the awesome work and have a good one! P.S. I’m still waiting for that juicy IMO playlist of problems...
The first part is like a deus ex machina... more systematic approach would be rewrite n=(ab-k) with k>=1 (ab)^2-4(a+b)=(ab-k)^2 (ab)^2-4(a+b)=(ab)^2-2kab+k^2 ; cancel (ab)^2, rearrange 2kab=4a+4b+k^2 ; 2ab
Ab+n=1or 2or 4 and ab+n=4(a+b) ,or 2(a+b) Or(a+b) respectively and since the equation is symmetric hence if (a,b) is a solution then (b,a) is also a solution
Ok, I just want to be sure I'm not misinterpreting this. In this case, you can assume a < or = b because in the original equation, the two terms are interchangeable, right?
@@琥珀-u3o It's a way of making the calculations easier. This way we just have to find the solutions for a =< b, and once we get these we instantly know the other solutions just by switching a and b, no need to calculate them.
Rather than applying case by case i solved like this. when 4(a+b)=>2ab-1, we can assume that 4(a+b)=>2ab because both terms are even number and we obtain 2a/(a-2)=>b and 2b/(b-2)=>a After that we multiply both sides and cancel out ab terms because it cannot be 0. At the end we can obtain 4=>(a-2)*(b-2).... After than this point we try all combinations.
Tried to do a,b∈Z instead of N, cases a=0 (b is perfect square), and a=-1 (always works) are easy. For a ≤ b < 0, we can use similar approach as for N (obtaining only b=-1 solution), but I see no nice way how to solve case a ≤ -2, 1≤ b, because |4(a+b)| may be arbitrarily small (giving trivial solution a=-b, but not ruling out other solutions).
Wait, 0 is not in N? Since when? Here in France we include 0 in N and usually note N* when we want only the positive integers. Is that a matter of country?
I think it's a matter of linguistics. People interpret the wording differently so I have two different university lectors with two different takes on it. If you mention that N includes 0, it shouldn't be faulted.
0 is in N when you need it and 0 is not in N when you need it. There’s no clear definition about that. It is a matter of country, it’s also a matter of context in a proof/theorem/exercise.
@@PlayerMathinson I had a discussion few days ago about “whole numbers”. For me it was ℤ, for others it was ℕ or even ℕ*. Another case of "to each their own".
I found out that for a,b more than or equal to 5, the given expression is between (ab)^2 and (ab-1)^2 , so no solutions there. Then the rest is casework !! Nice problem
I wonder if including that last solution would have been considered wrong by the Olympiad evaluators, since the problem specified that n was a natural number. At a minimum, I think one would have to qualify that solution with a conditional such as "If n can be 0...).
I converted this problem to "find two pairs of natural numbers such that the sum of one pair equals the product of the other pair" (see that 1+5=2x3 and 2+3=1x5). My solution is a little longer though. First, draw a square ABCD with side length ab. Then draw the diagonals. Draw a smaller square EFGH inside. the centroid of ABCD and EFGH must coincide. Now let the side length of EFGH be n. Notice that there are 4 trapezoids formed. Let the height of each of the trapezoid be h. Then we can see that n=ab-2h. The task will be to find h such that the area of one trapezoid is (a+b). Hence h(ab-h)=(a+b). We simplify to get the quadratic equation h^2-abh+(a+b)=0. Let's call this equation (1)Notice that the discriminant of this is the problem we are trying to solve. This means that the discriminant is a perfect square >0 and the roots will be real, which are (ab +/- n)/2. If both a and b are odd, then n is odd and so the roots are natural numbers. If either a or b is even, or both are even, then n is also even and h is still a natural number. Hence, we can denote the roots as natural numbers s,t. By inspection of equation (1), we see that st=(a+b) and s+t=ab. We can write another quadratic equation here where a and b are the roots. The quadratic equation is j^2-(st)j+(s+t)=0. Let us call this equation (2). Now we see that the determinant of equation (2) is d=(st)^2-4(s+t), and that both a and b are natural numbers and so this determinant is a perfect square also. This means that is we now a or b, we can use that to find another solution s and t using equation 1 and if we know a solution s and t using equation 2. Now consider first f(x,y)=xy/(x+y). Using calculus, we can see that f(x,y) is increasing for any natural numbers x and y. Notice also that when either x or y=1, then xy/(x+y)=2, y>=2, then f(x,y)>=1. Going back to a,b,s and t, we have st=a+b and s+t=ab (hence find two pairs of natural numbers such that the sum of one pair equals the product of the other pair). From this we have (st)/(s+t)=(a+b)/(ab). We see now that when the LHS >1 then the RHS 1, RHS1, which means that (ab)/(a+b)1, RHS
Nice video and problem, but if you're going to include 0 among the naturals, then (0,0) is a valid solution and you needed to take into account the case where a=0, which you didn't.
I'm not that bright; once figuring out the upper bound I was ready to just plug in everything from 0,0 to 4,4. Then midway through I remembered 0 isn't Natural
Let a+b=kl (i), ab=k+l (ii) where a,b,k,l are non negative integers. Then a^2b^2-4(a+b)=(k-l)^2. (i)-(ii) gives a+b-ab=kl-k-l => 2-(a-1)(b-1)=(k-1)(l-1)>=0 So, 2>= (a-1)(b-1)>=0 Suppose,(a-1)(b-1)=0 gives say a=1,then (k-1)(l-1)=2 => (k,l)=(3,2) or(2,3) and so (a,b)=(1,5) and similarly(a,b)=(5,1) By taking other two cases of (a-1)(b-1)=1,2 we get (a,b)=(2,3),(3,2),(2,2) solutions QED
Because if one integer is less than another, it's less than or equal to the previous integer, in general. The biggest integer that's less than n is n-1.
I assume so, the way I solved it was plotting (ab)^2 on a number line, and since n^2 has to be less than (ab)^2 the closest possible value of n is (ab-1)^2. I then drew a line from (ab)^2 towards 0 with a length of 4(a+b). After doing this, I observed that once (ab) is greater than a certain value, the line of length 4(a+b) will never reach the value (ab-1)^2 thus every value greater than or equal to that value of (ab) will not have a solution. I’m not sure if that’s considered a geometric proof, but I thought it was quite an intuitive observation.
i started with geometric representation then ended up changing the question to find two pairs of natural numbers where the sum of one pair is equal to the product of the other pair. not purely geometric as i only used that to set up the equations. then i used calculus and algebra.
for the case where n^2 + 8, and n^2 + 9, you could have noticed that the only square that is 8 less than another perfect square is n = 1, n^2 + 8 = 9, meaning (b - 2) = 3, b = 5; and for n^2 + 9 is a perfect square, that only works when n^2 = 16, or n^2 + 9 = 25, and then (2b - 1) = 5, so b = 3. In both cases it is easy to see that only one solution works, as the differences between consecutive squares are well known to be strictly increasing.
@@leif1075 actually these problems contain very creative and intuitive thinking, so usually such problem i like the most , so I got really excited to do this problem😂😂😂
I tried to do this question by myself. I did it this way: (ab)^2 - 4(a+b) = n^2 (ab)^2 - [2 * sqrt(a+b)]^2 = n^2 [ab - 2 * sqrt(a+b)] * [ab + 2 * sqrt(a+b)] = n^2 [ab - 2 * sqrt(a+b)] * [ab + 2 * sqrt(a+b)] = n * n u * v = n * n So I concluded that u = [ab - 2 * sqrt(a+b)] = n and v = [ab + 2 * sqrt(a+b)] = n So I concluded that u=v. (And that was bad assumption - I will write about this later below) u = v ab - 2 * sqrt(a+b) = ab + 2 * sqrt(a+b) -2 * sqrt(a+b) = 2 * sqrt(a+b) -4 * sqrt(a+b) = 0 sqrt(a+b) = 0 a+b = 0 b = -a In our original expression: (ab)^2 - 4(a+b) I substituted: b = -a and I get this: [a * (-a)]^2 - 4* [a + (-a)] [-a^2]^2 - 4* 0 a^4 - 0 a^4 And "a^4" of course is a perfect square since a^4 = (a^2)^2 But doing so I made a mistake. This is NOT so simple. I have made wrong assumption. I have overlooked such case when u is NOT equal to v but u*v IS a perfect squate, for example when u=4 and v=9, we get u*v = 4*9 = 36 = 6^2 so n=6 So we have the conclusion: Don't make the wrong assumptions. Don't overlook more complicated cases. Think: "Could there be some cases which I didn't think about?".
@Federico Rulli no ....one an maybe the only one is a=2 b=3 .... sorry but if don't have understand your opinion in totally .... obiusly for ab = a + b +1
An excellent rationale to designate successor-closed integer subsets by placing the starting value as subscript to the N (e.g., N₀ for the "natural" numbers, N₁ for the "counting" numbers, N₂ for the "plural" numbers, etc. ad nauseam).
@@TJStellmach Positive integers are positive integers. Natural numbers are natural numbers. #0isanatural. Anyhow, he uses 0 to construct one of the solutions...
Here's a fully deductive solution: First let's rule out the trivial solution (a,b;n) = (0,0;0). It can also be deduced but keeping a,b > 0 is helpful. (observe that if one of them is 0 and the other isn't then n² a+b (1) ab = a+b => a = b/(b-1) = 1+ 1/(b-1) => b=2 and a=2 this gives the second solution (2,2;0) (2) ab < a+b; by symmetry we can choose a >=b => ab < 2a so b k+m = a and km=a+1 => km = k+m+1, solving for k = (m+1)/(m-1) or k = 1+ 2/(m-1) => m = 2 and k=3 or m=3 and k=2 (an obvious symmetry) => a=5, b=1 this gives the third solution (5,1;1) (3) ab>a+b We don't need to investigate this case because by symmetry we can expect a=3 and b=2 (and k=5, m=1) (or you notice that in this case km < k+m and repeat the procedure in (2)) this gives the 4th solution (3,2;4) END PS: the first set of solutions can also be found by checking a=b => a^4 - 8a = n² => a(a-2)(a²+a+2)=n² if n=0 then a=0 or a=2 if n>0 => a|n and (a-2)|n => a²(a-2)²|n² => a(a-2)|(a²+a+2) => a|2 but this gives either 1 (no solution) or 2 (known solution)
Yes, you can at the very least simplify the sub-case analysis by nothing that a small value for a difference of squares is actually a very strict constraint. Several comments here go into this.
It is wrong in the results, because of the identity, to obtain values in the members of a and b , for it to be able to find those values, the identity had to have had an equality and that only works in equations of general function or of conservation function since there since it does not belong to one of the members of the identity there is no way to identify real elements in an identity, apart from ignoring the definition of the discriminant of the general equation of the second degree or also known as the general solution of factors, and by not considering it, this in turn modifies the identity, affecting all the real elements that belong to that identity and math is wrong, you should check it out
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Hi,
For fun:
1 "the next thing that we want to do",
1 "what I want to notice",
1 "so now what we are going to do",
2 "I'll go ahead and",
2 "so let's go ahead and",
1 "so let's may be go ahead and",
1 "I can go ahead and",
1 "we can go ahead and",
1 "and so on and so forth",
1 "great",
6:05 : a good place to look at.
6:31 : 2b or not 2b, that is the question.
19:48 : whouah! whouah !
A trick I found to make the solution quicker is that since n² has the same parity as a²b², we can use (ab - 2) as an upper bound for n, which leads to a more restrictive equation than with (ab - 1) and leaves us with only a handful of cases to check.
Good catch!
Can someone please explain Felissan solution? How is a^2b^2 parity same as n^2?
@@PlayerMathinson 4(a+b) is alway even. If a²b² is even the same is a²b²-4(a+b)=n². If a²b² is odd a²b²-4(a+b)=n² is odd. Thus they have the same parity
@@PlayerMathinson Also you could work mod 2. 4(a+b) vould vanish, leaving you with a²b²=n² (mod 2).
@@marcoantonioircide740 Thank you so much for making this clear.
From a former student of NUIG, Michael Tuite is a legend 😄
ab-2a-2b
Apparently my first comment did NOT appear for other people so let’s try again...
5:01 Good inequality to stop
6:01 Good place to look at
20:07 Sin áit mhaith le stopadh
Daily... How many elements do the following sets have :
A = {x ∈ ℚ | x = (n² + 1)/(2n² + n + 1), n ∈ ℕ, n ∈ [1, 1000]}
B = {x ∈ ℚ | x = (n² + 1)/(2n² + n + 1), n ∈ ℕ}
Posting hint in another comment.
Hint...
To be clear, we’re talking about sets, not multisets. So we look for distincts elements in the sets and you should look in what cases duplicates happen.
As none of the elements of A or B are zero we can consider their reciprocals minus 2. A reciprocal of a fraction as above can be written as 2+(n-1)/(n^2+1), so we have that the reciprocals minus 2 have the form (n-1)/(n^2+1), for appropriate values of n, or as m/(m^2+2m+2) for m=n-1. Taking another reciprocal (if m is not zero, thus n is not 1) and subtracting 2 yields m+2/m, whereas when m=0 the m/(m^2+2m+2)=0. So we need to see how many nonzero values m+2/m can assume when m lies in [0,999]. The function x+2/x is monotonically increasing for x>sqrt(2), so for m>=2 m+2/m will not be repeating values, and will increase for increasing m. When m=1 m+2/m=3, when m=2 it's also 3, so there is exactly one repeated value. Moreover 0 is never reached. Thus the cardinality of A is 1000-1=999. The same argument shows that the cardinality of B is countably infinite.
EDIT: I also need to remark that the operation of taking the reciprocal and subtracting 2 is injective. This is hopefully obvious/well known.
EDIT2: There is a very simple way to see that the cardinality of B is infinite while doing barely no work: Notice how the function (x^2+1)/(2x^2+x+1) is rational, and thus algebraic. We'll show that *No nonconstant algebraic functions can assume one value infinitely many times.* As a bonus we'll get that sin(x) is not algebraic :). The proof is extremely easy: say f(x) is algebraic, and p(x,f(x))=0 for some polynomial p. If f(x)=y for a fixed y then p(x,y)=0. As an equation of x, that is taking y as a fixed parameter, we can see that if it has infinitely many solutions then the polynomial must be constant in x. But then y must be constant. This concludes the proof of the mini-fact we'll use. Since no values can be attained infinitely many times the image of an infinite set must be infinite (of the same cardinality). This concludes the solution.
@@thephysicistcuber175 henlo lorenzo mauro, big fan
@@thephysicistcuber175 Forgot to come back here and clearly say that's the correct answer. I have nothing else to add, everything is in there :p
For common solution we must have n² + 1 and (n² + 1)+n² + n have a common factor. Hence n² + 1 and n² + n must have common factor which happens only fo n = 1. Hence 999 elements.
6:00 this can also be thought of as a rectangle with sides a and b, where the area is less than the perimeter! (without regard to units, that is.)
One could say he is a Math buff.
A buff math buff.
hes a rock climber
A rock star in math because even only 5% of math PROFESSORS(!) truly succeed in mathematical competitions like this. Studying maths in university is up to 1.5 orders of magnitude easier than this. Yes, you need to know much more than here but you need far less mathematical fantasy.
@@howmathematicianscreatemat9226 yeah, we don't have to use our brain this much to solve questions like these which require creativity & out of the box thinking sometimes. Sometimes understanding, remembering & regurgitating those truckloads of abstract theorems itself is a difficult task at university, let alone APPLYING them in creative ways to solve challenging problems 😅
@@howmathematicianscreatemat9226 It's up to 31.623 times easier?
I like this kind of problems.. When I watch the solution I feel like "That's easy and funny" but I can't do that alone
I feel the same lol
work and train hard you will get to improve and solve problem you once thought too hard for you
They are really freaking cool honestly, and there is a certain satisfaction in doubling back and making a full loop check even when you are sure of how you arrived at the answer. Finding that n is the same both in the algebra of the subcases and by substituting a and b from the subcase by which you arrived at an n into the original expresssion and coming up with that same n again is fundamentally sublime. You just kick back, and enjoy a nirvana of "Man, math just works" Not that I'm exceptional at it, but it goes w/o saying- math logic is remarkably better than any other because it's so fundamental.
There's a small shortcut for the a=2 case. Note that (2b-1)^2=n^2+3^2 is a Pythagorean triple with one side length of 3, implying n=4 and 2b-1=5. Although is does make it harder to realize the 0+9=9 case is actually relevant.
This is the best channel for math on youtube
06:03
a = 2.(a + b);
a b = a;
Substituting into the inequality::
a² >= 4.a; ==> a >= 4;
Hi Rogers. If you are interested in math competitions, please consider our channel ruclips.net/user/SQRTime. Hope to see you 😊
The only person who motivates me to become a mathematician!❤️
You mean be a buff mathematician
@@general9064 I mean a world-class mathematician,just like him😉
My thoughts how to solve this:
since (ab)^2 is already a square, we need to look at the difference between that and the perfect square just below it. Once a and b get beyond a certain since, this difference is going to be > 4(a+b) and therefore you only really need to test small values of a and b for that to hold.
When I solved this , I consider 0 NOT a Natural number. Natural numbers should be 1, 2, 3, .....
Why can you say n
case 1 has also (trivial) subcase 3: ((b-2)-n)((b-2)+n) = 8 may have both outer parenthesis to be negative. But that is possible only for b < 2, thus we should quickly check (and rule out) a=1, b=1.
Fun problem! But the parity argument (as mentioned below) plus some factoring significantly simplifies the cases to check. I also explored a=b in general which leads to the interesting fact that a^4-8a = a*(a-2)*(a^2+2a+4) is square ... but isn't much help in this problem but might be a useful approach for similar problems. (edited)
Why does ab
@@minime1235able That was a stupid error. Ugh! Thanks for the correction. My only excuse is that I approached the cases very differently and didn't put in the diligence when I followed his method.
I started with the parity argument to realize that n= n^2 you derive ab (b-2)^2 = n^2 +8
But 1 and 9 are the only perfect squares that differ by exactly eight [Recall that (k+2m)^2 - k^2 = 4km + 4m^2 = 4(k+m)m but this is only 8 if k=m=1]
Which yields (a,b)=(1,5) and its symmetric pair.
@@EdBailey1208 the other guy pointed out your mistake and you also agreed. So why don't you edit your original comment?
1. if a = 1 we have b^2-4*b-4 it has more than 1 root, thus can't be square but can be equal to 1 when x = 5. [1,5],[5,1]. 2 if a>=b>1 then a*b > a+ b or a+b = a*b-k with k>=1, we're getting x^2 - 4*x + 4*k which is always > 1 but can have one root when k = 1. so a*b = a+b-1 which can only be true in [2,3],[3,2]
case a=1: If (b-2)^2 = n^2+8 then you have 2 perfect squares which differ by 8. For (16,25) and above there are no solutions. So you look at all couples below that. And it's quickly seen that only (1,9) fits the bill.
Similarly for a=2, but this time since the difference is 9 you include(16,25) to be checked which gives a solution.
What do you mean? Where do you get 9 from if a equals 2 and where do yiu get b minus 2??
@@leif1075 Partway through manipulating b^2-4(b+1)=n^2 he gets (b-2)^2-n^2=8. But the difference of successive squares (to say nothing of more widely separated squares) grows quickly enough to rapidly eliminate the possibility of such a difference being as low as 8. At 5^2-4^2 it's already larger. Checking squares lower than 5^2 (i.e. 1, 4, 9, and 16) against each other, the only pair that differs by 8 is (1,9). So 9=(b-2)^2 in that solution.
Man your videos are really nice to watch
factorize
b^2·(a + 2·(√(1 - b^3) + 1)/b^2)·(a - 2·(√(1 - b^3) - 1)/b^2) = n^2
u can see for all values greater than b =1 , we are going to have complex number.
if b= 1 then (a+2)^2 = n^2 or a = n -2;
if b = 1 then b = n-2;
or as u said, a*b
lol my bad, is minus.. :(
Wow! Nice problem! I can’t really say that it was hard, but I have to admit that I enjoyed watching you explain it. Keep up the awesome work and have a good one!
P.S. I’m still waiting for that juicy IMO playlist of problems...
The dog was impressed at the end.
The first part is like a deus ex machina... more systematic approach would be
rewrite n=(ab-k) with k>=1
(ab)^2-4(a+b)=(ab-k)^2
(ab)^2-4(a+b)=(ab)^2-2kab+k^2 ; cancel (ab)^2, rearrange
2kab=4a+4b+k^2 ; 2ab
13:24 Shouldn't it be obvious from the +n and -n that the difference of the factors is 2n i.e. even so 1 and 8 will not work.
Ab+n=1or 2or 4 and ab+n=4(a+b) ,or 2(a+b) Or(a+b) respectively and since the equation is symmetric hence if (a,b) is a solution then (b,a) is also a solution
just switch to the point where you conclude that a has to be
3:50
I didn't get why we got the "ab-1" term
Awesome explanation.
The question domain for n is confusing. Why couldn't it just say non-negative integers?
Ok, I just want to be sure I'm not misinterpreting this. In this case, you can assume a < or = b because in the original equation, the two terms are interchangeable, right?
@@琥珀-u3o It's a way of making the calculations easier. This way we just have to find the solutions for a =< b, and once we get these we instantly know the other solutions just by switching a and b, no need to calculate them.
Rather than applying case by case i solved like this.
when 4(a+b)=>2ab-1, we can assume that 4(a+b)=>2ab because both terms are even number and we obtain 2a/(a-2)=>b and 2b/(b-2)=>a
After that we multiply both sides and cancel out ab terms because it cannot be 0.
At the end we can obtain 4=>(a-2)*(b-2).... After than this point we try all combinations.
Tried to do a,b∈Z instead of N, cases a=0 (b is perfect square), and a=-1 (always works) are easy. For a ≤ b < 0, we can use similar approach as for N (obtaining only b=-1 solution), but I see no nice way how to solve case a ≤ -2, 1≤ b, because |4(a+b)| may be arbitrarily small (giving trivial solution a=-b, but not ruling out other solutions).
Wait, 0 is not in N? Since when? Here in France we include 0 in N and usually note N* when we want only the positive integers. Is that a matter of country?
I had the same reaction, I checked and wikipedia says both definitions (with and without 0) are used...
I think it's a matter of linguistics. People interpret the wording differently so I have two different university lectors with two different takes on it. If you mention that N includes 0, it shouldn't be faulted.
0 is in N when you need it and 0 is not in N when you need it. There’s no clear definition about that.
It is a matter of country, it’s also a matter of context in a proof/theorem/exercise.
Yep in our country when you include 0 with N it is called whole numbers.
@@PlayerMathinson I had a discussion few days ago about “whole numbers”. For me it was ℤ, for others it was ℕ or even ℕ*.
Another case of "to each their own".
Quick solution: n^2
a+b 6
Typo in last line, it shall be a+b>=ab-1
20:07
I found out that for a,b more than or equal to 5, the given expression is between (ab)^2 and (ab-1)^2 , so no solutions there. Then the rest is casework !! Nice problem
I wonder if including that last solution would have been considered wrong by the Olympiad evaluators, since the problem specified that n was a natural number. At a minimum, I think one would have to qualify that solution with a conditional such as "If n can be 0...).
I converted this problem to "find two pairs of natural numbers such that the sum of one pair equals the product of the other pair" (see that 1+5=2x3 and 2+3=1x5). My solution is a little longer though. First, draw a square ABCD with side length ab. Then draw the diagonals. Draw a smaller square EFGH inside. the centroid of ABCD and EFGH must coincide. Now let the side length of EFGH be n. Notice that there are 4 trapezoids formed. Let the height of each of the trapezoid be h. Then we can see that n=ab-2h. The task will be to find h such that the area of one trapezoid is (a+b). Hence h(ab-h)=(a+b).
We simplify to get the quadratic equation h^2-abh+(a+b)=0. Let's call this equation (1)Notice that the discriminant of this is the problem we are trying to solve. This means that the discriminant is a perfect square >0 and the roots will be real, which are (ab +/- n)/2. If both a and b are odd, then n is odd and so the roots are natural numbers. If either a or b is even, or both are even, then n is also even and h is still a natural number. Hence, we can denote the roots as natural numbers s,t. By inspection of equation (1), we see that st=(a+b) and s+t=ab. We can write another quadratic equation here where a and b are the roots. The quadratic equation is j^2-(st)j+(s+t)=0. Let us call this equation (2). Now we see that the determinant of equation (2) is d=(st)^2-4(s+t), and that both a and b are natural numbers and so this determinant is a perfect square also. This means that is we now a or b, we can use that to find another solution s and t using equation 1 and if we know a solution s and t using equation 2.
Now consider first f(x,y)=xy/(x+y). Using calculus, we can see that f(x,y) is increasing for any natural numbers x and y. Notice also that when either x or y=1, then xy/(x+y)=2, y>=2, then f(x,y)>=1.
Going back to a,b,s and t, we have st=a+b and s+t=ab (hence find two pairs of natural numbers such that the sum of one pair equals the product of the other pair). From this we have (st)/(s+t)=(a+b)/(ab). We see now that when the LHS >1 then the RHS 1, RHS1, which means that (ab)/(a+b)1, RHS
That's awesome Solution!
Nice video and problem, but if you're going to include 0 among the naturals, then (0,0) is a valid solution and you needed to take into account the case where a=0, which you didn't.
I'm not that bright; once figuring out the upper bound I was ready to just plug in everything from 0,0 to 4,4. Then midway through I remembered 0 isn't Natural
Let a+b=kl (i), ab=k+l (ii) where a,b,k,l are non negative integers.
Then a^2b^2-4(a+b)=(k-l)^2.
(i)-(ii) gives a+b-ab=kl-k-l
=> 2-(a-1)(b-1)=(k-1)(l-1)>=0
So, 2>= (a-1)(b-1)>=0
Suppose,(a-1)(b-1)=0 gives say a=1,then (k-1)(l-1)=2 => (k,l)=(3,2) or(2,3) and so (a,b)=(1,5) and similarly(a,b)=(5,1)
By taking other two cases of
(a-1)(b-1)=1,2 we get (a,b)=(2,3),(3,2),(2,2) solutions
QED
In the last case of a=b=2, N=0, so not natural number?
For math competitions, please consider ruclips.net/video/LSypt8uSSW4/видео.html and other videos in the Olympiad playlist. Hope you enjoy 😊
Galway when pronounced sounds like "Gaulway" where Gaul sounds like the natives of what is now France.
He's definitely the Bob Ross of math !
Excuse me, in 2:51 . . . what was the justification for jumping from n < ab to n ≤ ab - 1 ?
That's how integers work. If a is integer and a
@@Milan_Openfeint
Oh, ok, I see. I've completely overlooked the 'n is an integer' part. Thanks!
2:49 Why is this true??
Because if one integer is less than another, it's less than or equal to the previous integer, in general. The biggest integer that's less than n is n-1.
I've seen this implication in other videos and I don't see why. If n
Keep in mind that ab is also an integer. So in the set of integers, a number x with ab-1
4:44 please explain why he simply dropped the one half
a and b are integers, so ab is an integer and so is 2a +2b. ab
@@fmakofmako Ah! I see. Thank you, mate
Im my country, 0 is a natural number
So case 3: a=0
-4b=n²
As -4b≤0 and n²≥0, b=0
Same
I was in Galway University a few weeks ago ... but just as a tourist.
9:37 oops does not work? this runs in milliseconds
10 for a=1 to 100:for b=1 to 100
20 dis=a*a*b*b:dis=dis-4*(a+b):if dis
Maybe we could use the sophie germain identity?
Wondering if there can be a geometric proof
That would be really cool. I’m sure one exists, only if someone was to go to the effort of finding it specific to this question. Unlikely
I assume so, the way I solved it was plotting (ab)^2 on a number line, and since n^2 has to be less than (ab)^2 the closest possible value of n is (ab-1)^2. I then drew a line from (ab)^2 towards 0 with a length of 4(a+b). After doing this, I observed that once (ab) is greater than a certain value, the line of length 4(a+b) will never reach the value (ab-1)^2 thus every value greater than or equal to that value of (ab) will not have a solution. I’m not sure if that’s considered a geometric proof, but I thought it was quite an intuitive observation.
i started with geometric representation then ended up changing the question to find two pairs of natural numbers where the sum of one pair is equal to the product of the other pair. not purely geometric as i only used that to set up the equations. then i used calculus and algebra.
if you assumed first n to be striclty positive, why not the second time?
His videos are awesome 😍 it inspires me to solve olympaids questions
He inspires me to get the gym more often. It's possible to be both smart and buff.
Isn't a=3, b=6 a solution for the equation ab=2a+2b too?
Applying congruences to the original equation, modulo 4, 6, and 8 might have been useful.
Why the couple (0;0) is not included ?
That pair has 0 and 0∉ℕ
Really intuitive and simple solution!
I believe for all of the system of equations we should also find n to check that it's integer
Keep up broo
Can you get us some abstract algebra medium problems .
In the first sight ,I can only get the solution (2.3) .Thanks for the other answer (1,5) .
When a=-b?
sir it can be solved by digital root method
for the case where n^2 + 8, and n^2 + 9, you could have noticed that the only square that is 8 less than another perfect square is n = 1, n^2 + 8 = 9, meaning (b - 2) = 3, b = 5; and for n^2 + 9 is a perfect square, that only works when n^2 = 16, or n^2 + 9 = 25, and then (2b - 1) = 5, so b = 3. In both cases it is easy to see that only one solution works, as the differences between consecutive squares are well known to be strictly increasing.
I love this problem , looks really exciting.
Why do you say exciting?
@@leif1075 actually these problems contain very creative and intuitive thinking, so usually such problem i like the most , so I got really excited to do this problem😂😂😂
@@leif1075 well from where are you bro?
BTW I am from INDIA
@@msdian3926 me 2 😍
These videos are the best
I tried to do this question by myself.
I did it this way:
(ab)^2 - 4(a+b) = n^2
(ab)^2 - [2 * sqrt(a+b)]^2 = n^2
[ab - 2 * sqrt(a+b)] * [ab + 2 * sqrt(a+b)] = n^2
[ab - 2 * sqrt(a+b)] * [ab + 2 * sqrt(a+b)] = n * n
u * v = n * n
So I concluded that
u = [ab - 2 * sqrt(a+b)] = n
and
v = [ab + 2 * sqrt(a+b)] = n
So I concluded that u=v.
(And that was bad assumption - I will write about this later below)
u = v
ab - 2 * sqrt(a+b) = ab + 2 * sqrt(a+b)
-2 * sqrt(a+b) = 2 * sqrt(a+b)
-4 * sqrt(a+b) = 0
sqrt(a+b) = 0
a+b = 0
b = -a
In our original expression:
(ab)^2 - 4(a+b)
I substituted:
b = -a
and I get this:
[a * (-a)]^2 - 4* [a + (-a)]
[-a^2]^2 - 4* 0
a^4 - 0
a^4
And "a^4" of course is a perfect square since
a^4 = (a^2)^2
But doing so I made a mistake.
This is NOT so simple.
I have made wrong assumption.
I have overlooked such case when u is NOT equal to v but u*v IS a perfect squate, for example when u=4 and v=9, we get
u*v = 4*9 = 36 = 6^2
so n=6
So we have the conclusion:
Don't make the wrong assumptions.
Don't overlook more complicated cases.
Think: "Could there be some cases which I didn't think about?".
when ab = a + b + 1 ? ... if is it correct ... i have spent 5 minutes to think about it
@Federico Rulli no ....one an maybe the only one is a=2 b=3 .... sorry but if don't have understand your opinion in totally .... obiusly for ab = a + b +1
Man wowwwww you are great i always learn from you
It would be interesting to see if the original question specifically allowed for n=0
Man i love math
Hello from Croatia. Which conference was it?
How did you guess that there must be only a few solutions to it?
I jumped in to the difference of squares and looked at factors of 4(a+b) and came to the same conclusions.
Thank you!
That a good channel to stop by ......😛😛
Great sir 😍loved your way
Best shoutout ever seen
0:11
Where is "Good place to stop"?
20:07
Hay thanks for the Question
LOVE FROM 🇮🇳DIA
An excellent rationale to designate successor-closed integer subsets by placing the starting value as subscript to the N (e.g., N₀ for the "natural" numbers, N₁ for the "counting" numbers, N₂ for the "plural" numbers, etc. ad nauseam).
9:50 Why not just say "144-28=116, which is not a square"?
Why perform the multiplication when you don't need to? (To get 144 and 28)
@@shen144 Recalling facts from your times tables is not "performing multiplication".
I don't know, but I would do it the same way as Michael.
You forgot a=0,b=0 if you include this it also works
Can be solve easily by using discriminant of quadratic function.
The dog is against considering 0 as an option for n
Professor,
This exercise have only 3 solutions ((1,5);(2,2) and (2,3)), b/c you prove in the beggining a≤2 ( a=1 or a=2)
Really nice. But you forgot a=b=0. ;-)
No, because the question calls for a and b to be natural numbers (i.e. positive integers).
@@TJStellmach Positive integers are positive integers. Natural numbers are natural numbers. #0isanatural.
Anyhow, he uses 0 to construct one of the solutions...
Here's a fully deductive solution:
First let's rule out the trivial solution (a,b;n) = (0,0;0). It can also be deduced but keeping a,b > 0 is helpful.
(observe that if one of them is 0 and the other isn't then n² a+b
(1) ab = a+b => a = b/(b-1) = 1+ 1/(b-1) => b=2 and a=2
this gives the second solution (2,2;0)
(2) ab < a+b;
by symmetry we can choose a >=b
=> ab < 2a so b k+m = a and km=a+1
=> km = k+m+1, solving for k = (m+1)/(m-1) or k = 1+ 2/(m-1)
=> m = 2 and k=3 or m=3 and k=2 (an obvious symmetry)
=> a=5, b=1
this gives the third solution (5,1;1)
(3) ab>a+b
We don't need to investigate this case because by symmetry we can expect a=3 and b=2 (and k=5, m=1)
(or you notice that in this case km < k+m and repeat the procedure in (2))
this gives the 4th solution (3,2;4)
END
PS: the first set of solutions can also be found by checking a=b
=> a^4 - 8a = n² => a(a-2)(a²+a+2)=n²
if n=0 then a=0 or a=2
if n>0
=> a|n and (a-2)|n => a²(a-2)²|n² => a(a-2)|(a²+a+2) => a|2 but this gives either 1 (no solution) or 2 (known solution)
Quite tedious way to solve that; isn’t there any faster way ?
Yes, you can at the very least simplify the sub-case analysis by nothing that a small value for a difference of squares is actually a very strict constraint. Several comments here go into this.
n cannot be zero as per the initial conditions
You're gonna have to post dog photos in the community tab now. ;p
(a+b)^2-4ab=n^2 always sir. can we say that a+b =ab and it will always be true for given equation also😂😂😂 sorry just joking
In my comment, I have written the solution in your way.
Y not just a = -b? I didn't watch the video, but I was just intrigued by the thumbnail
natural numbers...
It is wrong in the results, because of the identity, to obtain values in the members of a and b , for it to be able to find those values, the identity had to have had an equality and that only works in equations of general function or of conservation function since there since it does not belong to one of the members of the identity there is no way to identify real elements in an identity, apart from ignoring the definition of the discriminant of the general equation of the second degree or also known as the general solution of factors, and by not considering it, this in turn modifies the identity, affecting all the real elements that belong to that identity and math is wrong, you should check it out
Nice!
I’m from Ireland. Really cool to see this.
The way he says Galway though
Im very bad at math. But it feels so cool to watch this😎
i am Chinese ，I also solved it, the method is a bit different