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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?

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  • $\begingroup$ 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. $\endgroup$ – topoquestion Feb 25 '16 at 21:11
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Feb 25 '16 at 21:35
  • $\begingroup$ 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 :-( $\endgroup$ – Jyrki Lahtonen Feb 25 '16 at 22:07
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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 $(m,n)=(1,-1),(1,0)$.

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  • $\begingroup$ Very nicely done. Thank you. Unless there is a simpler approach, this may be another exercise that Burton made harder than intended - imaginary numbers used in your linked paper are not even introduced in this textbook. I am using a very early edition since I found a cheap hardcopy. At any rate, I learned about Mordell equations. $\endgroup$ – topoquestion Feb 26 '16 at 1:09
  • $\begingroup$ @topoquestion Where exactly did you find this problem? $\endgroup$ – user236182 Feb 26 '16 at 1:15
  • $\begingroup$ Elementary Number Theory Revised Printing. Copyright 1980, 1976. So 2nd edition perhaps. By David M. Burton. This is exercise 17 in section 12.2 I don't know, can I unaccept an answer? $\endgroup$ – topoquestion Feb 26 '16 at 1:18
  • $\begingroup$ @topoquestion You've just unaccepted it. $\endgroup$ – user236182 Feb 26 '16 at 1:20

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