Solving m^2=n^3+k^2 (partially) for Positive Integers

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[SyberMath]

Hello everyone,
I did not find all the solutions. As some textbooks say…The rest is left as an exercise to the reader…🥸🤪🙃

[MrLidless]

Using the sum of cubes formula gives n³ = (n(n+1)/2)² - (n(n-1)/2)², so there is at least one solution for every n>1.

[DrWeselcouch]

At first I thought the power of m and k was 3 and I was nervous that you found a counter example to Fermat's Last Theorem.

[armacham]

m^2 = n^3 + k^2
subtract k^2 from both sides
m^2 - k^2 = n^3
(m - k)(m + k) = n^3

this can be true when:
m-k = 1 and m+k = n^3
m-k = n and m+k = n^2

but when n is not a prime number, there are even more solutions possible
e.g. n=8, m-k=4, m+k=128.

m-k can be less than n, or greater than n, but obviously it can't be greater than n*sqrtn

if n = a*b*c, then you could have

m-k = 1, m+k = a^3b^3c^3
m-k = a, m+k = a^2b^3c^3
m-k = b, m+k = a^3b^2c^3
m-k = c, m+k = a^3b^3c^2
m-k = ab, m+k = a^2b^2c^3
m-k = ac, m+k = a^2b^3c^2
m-k = bc, m+k = a^3b^2c^2
m-k = abc, m+k = a^2b^2c^2
m-k = aa, m+k = ab^3c^3
m-k = abca, m+k = ab^2c^2
etc.etc.etc.
it seems like there are too many possibilities for there to be a general solution . . . at least, from the way I approached the problem. I'm expecting the video will show a radically different approach

[davidblauyoutube]

The most general solution that I could find is (assuming these expressions are all positive integers, which they are under various conditions that I won't list here):
m = h z ( u^3 z + h y^3 ) / 2
n = h z ( u y )
k = h z ( u^3 z - h y^3 ) / 2

My solution relied heavily on the use of GCD.

[leickrobinson]

The 2nd method can be generalized to find other solutions, by considering other factorizations of n³. (even if n is prime, the factor pair {1, n³} can yield an additional solution.)
For example, (17, 4, 15) is also a solution.
(However, only factor pairs with the same parity will work. Why that is so I will leave as “homework“.) :-D

[JohnRandomness]

Give me a challenge, I like to try it myself. I'm not quite spoiling the video, because I haven't proven that this pattern exhausts the solutions.

1 = 1 + 0 -- outside the requirements, but helps the pattern.
9 = 8 + 1 -- n can be any positive integer except 1.
36 = 27 + 9 -- m: add n to the previous m, or (sum of 1 to n) = n(n+1)/2
100 = 64 + 36 -- k: (previous m) = n(n-1)/2
225 = 125 + 100 -- Prove that m² - k² = n³.

m² - k² = (m+k)(m-k) = [ n(n+1) + n(n-1)][ n(n+1) - n(n-1)]/4 = (2n²)(2n)/4 = n³

[ivankaznacheyeu]

It's easy to express general solution from n^3=(m-k)(m+k). m=(a+b)/2, k=(a-b)/2, where a is any integer divisor of n^3 that is greater than n^(3/2), such that b=n^3/a has the same parity as a. a=n^2 works for any integer n > 1.

[ompatil]

I found a more general solution for even n
(m, n, k) = (2t^3/s + s, 2t, 2t^3/s - s)
where t is any integer and s is any integer that divides 2t^3

[sanjuiyengar]

another observation in this problem is that for any given n, the maximum possible acceptable difference between m and k that satisfy the given equation is n, ie, m-k=n even though other possibilities exist

[thierryvieville]

That an interesting problem, thank you, very beginner regarding diophantine equations, i tried to investigate a bit the video 2nd method :
m^2 = n^3 + k^2 <=> p q = n^3, m = (p + q) / 2, k = (p - q) / 2
thus p = m + k and q = m - k must have the same parity [1] with p >= q, while m >= k.

For any n > 0 (see [2] for n = 0), we thus obtain, thanks to [3]:
p = n^2, q = n have same parity because of [3] and ((n^2+n)/2, n, (n^2-n)/2) are solutions as stated in the video,
p = n^3, q = 1 if and only n is odd because of [3] and ((n^3+1)/2, n, (n^3-1)/2) are solutions as mentionned by Leick Robinson
and, as Jorge Lozano Alcaido was thinking of if n is prime these are the only solutions, since these are the only decompositions.

This proves that there are solutions for all n, while [2] shows it is the case for m and k also.

Based on @SyberMath 2nd method a quadratic algorithm generates all solutions for say n < N
For any 0 < p <= N^3
For any 0 < q <= p with p q < N^3
If p and q have the same arity and p q is a cube
((p+q)/2, n, (p-q)/2)

[1, 1, 0], [3, 2, 1], [6, 3, 3], [14, 3, 13], [8, 4, 0], [10, 4, 6], [17, 4, 15], [15, 5, 10], [63, 5, 62], [15, 6, 3], [21, 6, 15], [29, 6, 25], [55, 6, 53], [28, 7, 21], [172, 7, 171], [24, 8, 8], [36, 8, 28], [66, 8, 62], [129, 8, 127], [27, 9, 0], [45, 9, 36], [123, 9, 120], [365, 9, 364], [35, 10, 15], [55, 10, 45], [127, 10, 123], [251, 10, 249], ...

I could not find a better way to enumerate all solutions ... could it be ?

[1] Same parity means either odd or even together.
[2] If n = 0 any m = k is a solution, showing that there are solutions for all k and m.
[3] The fact that n and n^d, d > 0 have the same parityi s easily obtained writing:
(2 n)^d= 2 (2^(d-1) n^d)
(2 n + 1)^d = 1 + 2 sum( (n p) 2^(p-1) n^p, p = 1..d) from the binomial formula.

[oguzhanozdogan]

2.yöntemde, n^3=m^2-k^2 eşitliğinde, m-k=1, m+k=n^3 ya da m-k=n, m+k=n^2 olma zorunluluğu yoktur. Örneğin n=5 iken m=63 ve n=62 olabilir. Ki bu bulmuş olduğunuz formülde çalışmıyor :)
Ancak 1.yönteminiz çok iyi!

[JohnRandomness]

Suppose k = m - 3. Then m² - k² = 6m - 9 = n³. n³ is a multiple of 3, so n is a multiple of 3 and n³ is a multiple of 3³. m = (n³ + 9)/6. Suppose n = 3. n³ = 27. m = 6 and k = 3. (6, 3, 3) is the same solution as SyberMath's. k = m - 1 gives (14, 3, 13).

n = 6: n³ = 216. m = 225/6 doesn't work. n has to be an *odd* multiple of 3.

n = 9: n³ = 729. m = 738/6 = 123, and k = 120. (123, 9, 120) accompanying (45, 9, 36) and (365, 9, 364).

[sanjuiyengar]

General solution for m, k odd and n even is m=(n^3-4)/4 and k=(n^3+4)/4
General solution for m, k even and n even if m=(n^3-16)/8 and k=(n^3+16)/8 but with the condition that n is of the form 4p where p is a natural no even or odd for even n of the form 2p, m and k are odd
For n odd m is odd and k is even or vice versa in this case they are consecutive integers m=(n^3-1)/2 and k=(n^3+1)/2 this is a more complete solution since this will give both m=62 and k=66 and for n=8 as well as m=127 and k=129

[jorgelozanoalcaido]

I love your videos!!!!
I reaaly love it but I have a doubt. What happens if n is not a prime number. For example imagine that n is a two prime numbers product (p and q for example). You have 12 different cases in the equation (m-k)(m+k)=(pq)^3. So imagine all the different cases that you can have with n a not prime number (different to 1 obviously). You will hace probably infinite solution. What do you think?

[WahranRai]

Strange result in the second method if Z+ doesnt include 0 !
For n=1 k= (n^2-n) /2 = 0

[jamescollis]

If you're not finding all solutions, you're not 'solving' the problem.

[JohnRandomness]

I saw someone claim a 2nd solution: (66, 8, 62). 66² - 62² = 8³? Yes. The solution I got was (36, 8, 28). Any pattern there? I notice that in one case, the difference between m and k is 4, while in the other case, it's 8.

EDIT: If n is odd, there's always n³ = 2m -1 ==> m = (n³ + 1)/2 and k = m - 1 = (n³ - 1)/2.
m² - k² = (n^6 + 2n³ + 1 - n^6 + 2n³ - 1)/4 = n³

EDIT: If n is even, try n³ = 4m - 4, k = m - 2. Then m² - k² = m² - (m - 2)² = 4m - 4 = n³. So if n=8, then m=129 and k = 127. Another solution is (129, 8, 127).