Integer solutions to cubic equation

Question

Part of something I'm doing at the moment results in the equation

$3k^2 + 3k + 1 = w^3$

where $k$ and $w$ are positive integers. From inspection it seems that the only integer solutions to this are the trivial pairs $(k,w) = (0,1)$, $(k,w)=(-1,1)$. This is fine, but I think I should be able to prove that these are the only possible solutions. I do not however know how to proceed - I note that the above equation can be written as

$6M + 1 = w^3$

where $M \geq 0$. Manipulating this identity, I can show that $w = 6Q +1$ but I'm not sure that helps me. Any ideas how once can proceed with this kind of proof, or is there an obvious reason why only the trivial solutions can satisfy the equation?

Apologies, typo corrected - k and w are both unknowns, I am simply trying to find all the pairs of integers that satisfy them. The bracketed terms are the solution sets. I've edited these to be clearer now... — DRG, Jan 09 '19 at 23:00
Yes - any integer again. Not sure how useful it is, but put it in for clarity. — DRG, Jan 09 '19 at 23:03
After minor transformations $k\leftarrow \frac{y-1}2$, you are looking for integer points on $$3y^2=4w^3-1$$ (in fact, you want $y$ odd). — Hagen von Eitzen, Jan 09 '19 at 23:16
@HagenvonEitzen: and in fact one has no choice or will in the matter of $y$ odd, since $4w^2$ is even etc. etc. etc. — Robert Lewis, Jan 11 '19 at 07:45

score 6 · Answer 1 · answered Jan 09 '19 at 23:16

6

The equation is equivalent to $$w^3+k^3=(k+1)^3\ .$$ This has no integer solutions except when one of the terms is zero, which gives precisely the solutions you have found.

Theorem (Euler?). The equation $x^3+y^3=z^3$ has no solutions in non-zero integers.

answered Jan 09 '19 at 23:16

David

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Good observation - you're right, I think that's Fermat's last theorem, but I'm curious if there's other proofs, because I have a few other forms I'd like to test for the existence of a cubic root! – DRG Jan 09 '19 at 23:37
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It's a special case of FLT... or actually, a special case of a special case of FLT :) – David Jan 09 '19 at 23:42
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The case $p=3$ was first proved by Euler, I think. – David Jan 09 '19 at 23:42

Integer solutions to cubic equation

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