# $n^2 + (n+1)^2 = m^3$ has no solution in the positive integers

The problem from Burton: show that the equation $n^2 + (n+1)^2 = m^3$ has no solution in the positive integers.

So far, I can see that gcd($n$,$n+1$)$=1$ and $m \equiv_4 1$ and $m=a^2 + b^2$ for some integers a,b. I'm guessing I need to reach a contradiction.

At this point, I am stuck. Any hints?

• Thanks Ivan, that is a good observation. I updated title. Another poster commented and deleted an approach using quadratic formula. I think he made a mistake in his c algebra but this leads to showing $2m^3 + 1$ must be a square if there is a solution. That's another angle. – topoquestion Feb 25 '16 at 21:11
• There was another mistake in the deleted answer in that the denominator should be $4$, not $2$. That makes the whole thing not work. It might be possible to rescue. – Ross Millikan Feb 25 '16 at 21:35
• Why do I get the vibe that André Nicolas posted a nice answer to this years ago? In other words, this is probably a duplicate. Can't find it now :-( – Jyrki Lahtonen Feb 25 '16 at 22:07

The equation is equivalent to $(2(2n+1))^2+4=(2m)^3$.
$a^2+4=b^3$ with $a,b\in\mathbb Z$ is a Mordell's Equation. See this paper (page $6$) for an elementary solution that uses $\Bbb Z[i]$ (the Gaussian integers). The only solutions are $(a,b)=(\pm 2,2), (\pm 11, 5)$.
$2m$ is even, so $2m=2$ and $2(2n+1)=\pm 2$, so the only integer solutions are $(m,n)=(1,-1),(1,0)$.