Alternative solution: Firstly all variables that appear will be positive integers Secondly some of the equal signs represents the ‘triple bar’ symbol so a=b(mod c) means a is congruent to b modulo c Since n^3 + 1 = 1(mod n) and mn - 1 = -1(mod n), (n^3 + 1)/(mn - 1) = -1 (mod n) So let n^3 + 1 = (mn-1)(kn-1) for some k Expanding and simplifying gives a quadratic in n: n^2 - mkn + (m+k) = 0 Since n is a positive integer, it is rational so the discriminant must be a perfect square So (mk)^2 -4(m+k) is a perfect square and w.l.o.g. let m >= k Obviously the expression is less than (mk)^2 so for it to be a perfect square it must be smaller or equal to (mk-1)^2 Expanding and rearranging gives 4m + 4k >= 2mk - 1 For k = 1, m>=1 For k=2, m>=2 For k=3,6>=m>=3 For k=4, m=4 For k>=5, no solution For k=3,4, we can check each of the possible pairs of (k,m) to see whether n is an integer. There are no solutions. For k=1, we have (n^3 + 1) is divisible by (n-1), so (n^3 + 1)/(n-1) = n^2 + n + 1 + 2/(n-1), so 2/(n-1) is an integer, n = 2 or 3. (The w.l.o.g. m>=k was just an assumption to bound the (m,k) pairs to find n. Now that the n can be found, we cannot hold this assumption anymore or else we will miss some solutions) For n=2, the factors of (n^3 + 1)=9 in the form of 2m-1 are 1,3,9, so m=1,2,5 respectively For n=3, the factors of (n^3 + 1)=28 in the form of 3m-1 are 2 and 14, corresponding to m=1 and m=5 For k=2, the discriminant 4m^2 - 4m - 8 is a perfect square, and denote it by c^2. We have c^2 + 3^2 = (2m - 1)^2, we can either rearrange c to the rhs and factor both sides to solve that c = 0 and m = 2 or c = 4 and m = 3 Finally we solve for n^3 + 1 = (2n-1)(2n-1) and n^3 + 1 = (3n-1)(2n-1) respectively, the first one giving n = 2 which we have already considered, and n=1 or n=5 for the latter equation For n = 1, n^3 + 1 = 2, so m = 2 or 3 (obviously) For n = 5, n^3 + 1 = 126. The factors of 126 are 1,2,7,14,3,6,21,42,9,18,63,126 (sorry for the non-ascending order), of which 9 and 14 are in the form of 5m - 1, so m = 2 or 3 Cheng Nick Hang: Concluding all results we have (n,m) = (1,2),(1,3)(2,1)(2,2),(2,5)(3,1)(3,5),(5,2),(5,3) It can be checked that each solution satisfies the given condition.
Good evening! What I liked more is that the quadratic equation. To find numbers that the sum is equal to the product of k and n and their product is the sum of k and n.
Let (n^3 + 1)/(mn-1) be x, so that (mn-1)x = n^3 + 1. Modulo n, we get (-1)x = 1, so that x = -1 (mod n) and hence x = kn-1 for some integer k. Then, n^3+1 = (kn-1)(mn-1). Expanding, n^3+1 = kmn^2 - (k+m)n + 1, or n^2 = kmn - (k+m). Thus k+m is a multiple of n, say k+m = jn. So n^2 = kmn - jn, or j+n = km. So we have that k+m = jn and j+n = km for integers j, k, m, and n. By symmetry, we can swap m and k; j and n; and (m,k) and (n,j). As such, if we let A = k+m = jn and B = j+n = km, WLOG A >= B, so k+m >= km. Rearranging, 1 >= km-k-m+1 = (k-1)(m-1). Again WLOG k >= m. If m = 1, the inequality holds, and if k = m = 2, the inequality holds; otherwise, the right-hand side is too large. We can investigate each of these scenarios case-by-case. Case 1: m = k = 2. Then, j+n = km = 4 and jn = k+m = 4. This means j and n are the roots of x^2 - 4x + 4 = 0, so j = n = 2. In this case, (m,n,j,k) = (2,2,2,2). Relaxing our generalities by allowing k and m to swap, j and n to swap, or (m,k) and (n,j) to swap does not give us any more solutions. Case 2: m = 1. Then j+n = k and jn = k+1, so jn = j+n+1 ; jn - j - n + 1 = 2 ; and (j-1)(n-1) = 2. j-1 and n-1 are both non-negative integers, so they're 1 and 2 in some order, so that j and n are 2 and 3. Then, k = 5. So we get (m,n,j,k) = (1,2,3,5) or (1,3,2,5). Relaxing our generalities, we also get (m,n,j,k) = (5,2,3,1), (5,3,2,1), (2,1,5,3), (2,5,1,3), (3,1,5,2), and (3,5,1,2). So we have a list of possible solutions for (m,n) as: (1,2), (1,3), (2,1), (2,2), (2,5), (3,1), (3,5), (5,2) and (5,3). By inspection, all nine of these are solutions, and so this is the full list of solutions.
As seems to be the case with your videos, I get about half way through on my own then I need to look at your solution. I wondered about providing by induction that n>=3 has no solutions with n=3 as the base case. But that just leaves us with trial and error for n=1 and n=2.
you should have shown that k>=1 since kn-1=+ve integer.. since (n^3+1)/(mn-1) is never zero or -ve for n,m∈Natural number.. => kn-1>=1 => kn>=2..=> n>=1.. now the inequality (k-1)n
can someone explain to me what's that assumption after "without lost of generality" ? he assumed m bigger than n why for example didn't he assume m smaller than n , what would happen then ?
yeah, after proving symmetry deciding which variable you choose to be biggest is up to you, but usually lots of people assume n>m (which is kinda strange, 'cause in alphabet m takes place before n)
Since whenever (m,n) is a solution, (n,m) is also a solution, we can first find all solutions where m >= n, and then to find the ones where n >= m we just swap m and n in our original list of solutions Whenever you have a symmetry like this, it allows you to impose a condition on the relative status of your variables, because you can recover the other solutions by applying the symmetry to the solutions you found. As long as every solution is in the same equivalence class as a solution that obeys your condition, you can assume your condition Without Loss of Generality.
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Alternative solution:
Firstly all variables that appear will be positive integers
Secondly some of the equal signs represents the ‘triple bar’ symbol so a=b(mod c) means a is congruent to b modulo c
Since n^3 + 1 = 1(mod n) and mn - 1 = -1(mod n), (n^3 + 1)/(mn - 1) = -1 (mod n)
So let n^3 + 1 = (mn-1)(kn-1) for some k
Expanding and simplifying gives a quadratic in n:
n^2 - mkn + (m+k) = 0
Since n is a positive integer, it is rational so the discriminant must be a perfect square
So (mk)^2 -4(m+k) is a perfect square and w.l.o.g. let m >= k
Obviously the expression is less than (mk)^2 so for it to be a perfect square it must be smaller or equal to (mk-1)^2
Expanding and rearranging gives 4m + 4k >= 2mk - 1
For k = 1, m>=1
For k=2, m>=2
For k=3,6>=m>=3
For k=4, m=4
For k>=5, no solution
For k=3,4, we can check each of the possible pairs of (k,m) to see whether n is an integer.
There are no solutions.
For k=1, we have (n^3 + 1) is divisible by (n-1), so (n^3 + 1)/(n-1) = n^2 + n + 1 + 2/(n-1), so 2/(n-1) is an integer, n = 2 or 3.
(The w.l.o.g. m>=k was just an assumption to bound the (m,k) pairs to find n. Now that the n can be found, we cannot hold this assumption anymore or else we will miss some solutions)
For n=2, the factors of (n^3 + 1)=9 in the form of 2m-1 are 1,3,9, so m=1,2,5 respectively
For n=3, the factors of (n^3 + 1)=28 in the form of 3m-1 are 2 and 14, corresponding to m=1 and m=5
For k=2, the discriminant 4m^2 - 4m - 8 is a perfect square, and denote it by c^2.
We have c^2 + 3^2 = (2m - 1)^2, we can either rearrange c to the rhs and factor both sides to solve that c = 0 and m = 2 or c = 4 and m = 3
Finally we solve for n^3 + 1 = (2n-1)(2n-1) and n^3 + 1 = (3n-1)(2n-1) respectively, the first one giving n = 2 which we have already considered, and n=1 or n=5 for the latter equation
For n = 1, n^3 + 1 = 2, so m = 2 or 3 (obviously)
For n = 5, n^3 + 1 = 126. The factors of 126 are 1,2,7,14,3,6,21,42,9,18,63,126 (sorry for the non-ascending order), of which 9 and 14 are in the form of 5m - 1, so m = 2 or 3
Cheng Nick Hang: Concluding all results we have (n,m) = (1,2),(1,3)(2,1)(2,2),(2,5)(3,1)(3,5),(5,2),(5,3)
It can be checked that each solution satisfies the given condition.
Good evening! What I liked more is that the quadratic equation. To find numbers that the sum is equal to the product of k and n and their product is the sum of k and n.
I also went down the mod n route but you actually completed it whereas I gave up!
Hi
Why no solution for k greater than or equal to 5?
Thanks
Let (n^3 + 1)/(mn-1) be x, so that (mn-1)x = n^3 + 1. Modulo n, we get (-1)x = 1, so that x = -1 (mod n) and hence x = kn-1 for some integer k.
Then, n^3+1 = (kn-1)(mn-1). Expanding, n^3+1 = kmn^2 - (k+m)n + 1, or n^2 = kmn - (k+m). Thus k+m is a multiple of n, say k+m = jn.
So n^2 = kmn - jn, or j+n = km.
So we have that k+m = jn and j+n = km for integers j, k, m, and n. By symmetry, we can swap m and k; j and n; and (m,k) and (n,j). As such, if we let A = k+m = jn and B = j+n = km, WLOG A >= B, so k+m >= km. Rearranging, 1 >= km-k-m+1 = (k-1)(m-1). Again WLOG k >= m. If m = 1, the inequality holds, and if k = m = 2, the inequality holds; otherwise, the right-hand side is too large. We can investigate each of these scenarios case-by-case.
Case 1: m = k = 2. Then, j+n = km = 4 and jn = k+m = 4. This means j and n are the roots of x^2 - 4x + 4 = 0, so j = n = 2. In this case, (m,n,j,k) = (2,2,2,2). Relaxing our generalities by allowing k and m to swap, j and n to swap, or (m,k) and (n,j) to swap does not give us any more solutions.
Case 2: m = 1. Then j+n = k and jn = k+1, so jn = j+n+1 ; jn - j - n + 1 = 2 ; and (j-1)(n-1) = 2. j-1 and n-1 are both non-negative integers, so they're 1 and 2 in some order, so that j and n are 2 and 3. Then, k = 5. So we get (m,n,j,k) = (1,2,3,5) or (1,3,2,5). Relaxing our generalities, we also get (m,n,j,k) = (5,2,3,1), (5,3,2,1), (2,1,5,3), (2,5,1,3), (3,1,5,2), and (3,5,1,2).
So we have a list of possible solutions for (m,n) as: (1,2), (1,3), (2,1), (2,2), (2,5), (3,1), (3,5), (5,2) and (5,3). By inspection, all nine of these are solutions, and so this is the full list of solutions.
Great Solution for Great Problem👍👍
Amazing method🔥 I was having a hard time solving it. Should've considered the inequalities rather than focusing too much on GCDs.
Yeah! That inequality for m>1 , and using the WLOG argument, so brilliant...
@@adamswilliam3462 so true
@@Goku_is_my_idol lmao
Hi! What type of mathematics is this?! I want to learn it!!! It looks so amazing! Any books you recommend please?
number theory
@@harshsinghal5898 thank you 🙏🏼 👍
Kindly raise the volume of your Microphone it’s kinda low
Btw great solution
Good evening! Very beautiful solution.
As seems to be the case with your videos, I get about half way through on my own then I need to look at your solution.
I wondered about providing by induction that n>=3 has no solutions with n=3 as the base case. But that just leaves us with trial and error for n=1 and n=2.
you should have shown that k>=1 since kn-1=+ve integer.. since (n^3+1)/(mn-1) is never zero or -ve for n,m∈Natural number.. => kn-1>=1 => kn>=2..=> n>=1.. now the inequality (k-1)n
I wonder why you don't consider the case n=0. For n=0 then for any number m the fraction is always an integer (=-1)
0 is not natural nymber
@@ЕгорСопожников , good evening!
It is intersting, in Brazil we consider 0 as a natural number.
Superb!
Nice)
great method !!!
bro is insane
Top-notch solution, sir. Thank you very much.
can someone explain to me what's that assumption after "without lost of generality" ? he assumed m bigger than n
why for example didn't he assume m smaller than n , what would happen then ?
yeah, after proving symmetry deciding which variable you choose to be biggest is up to you, but usually lots of people assume n>m (which is kinda strange, 'cause in alphabet m takes place before n)
Since whenever (m,n) is a solution, (n,m) is also a solution, we can first find all solutions where m >= n, and then to find the ones where n >= m we just swap m and n in our original list of solutions
Whenever you have a symmetry like this, it allows you to impose a condition on the relative status of your variables, because you can recover the other solutions by applying the symmetry to the solutions you found. As long as every solution is in the same equivalence class as a solution that obeys your condition, you can assume your condition Without Loss of Generality.
Damn boy
m=2, n=2
asnwer=1try study take pote
Riemann hypothesis has been solved
were lost🤣🤣🤣🤣😪😪😪😪
I solved it within 3 minutes
Alhamdulillah
Not beautiful solution
Life isn’t always beautiful, yknow. In mathematics, it just has to work. Being beautiful is a bonus
u have strong asian accent
he's from hongkong so reasonable
my math teacher also has this kind of accent. It is Hong Kong accent more than Asian accent