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The question: "Prove that there are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers." How would you go about picking a method to prove this? Is there a better way to do it other than by exhaustion?

nucleic
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    Fermat's last theorem.. done? – Euler....IS_ALIVE Feb 11 '13 at 05:14
  • It seems likely that exhaustion is what they want. This is the first and easiest case of Fermat's Last Theorem. There is a proof for all numbers (meaning bigger than 1000 is also allowed) that is often done in undergraduate number theory classes. But I think it's a bit long for this. Maybe you can do something by factoring the sum of two cubes? Perhaps you can rule it out when all three numbers (to be cubed) are below 10. – Will Jagy Feb 11 '13 at 05:14
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    There are very few perfect cubes in your range. And for any perfect cube, like $512$, one of the component cubes would have to be greater than $512/2$. The only candidate is $343$, and the other one would have to be odd, and therefore $125$ or less, bad! – André Nicolas Feb 11 '13 at 05:20
  • Interesting, thanks – nucleic Feb 11 '13 at 05:57

2 Answers2

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There are only $9$ perfect cubes less than $1000$, so there are not many cases to check. It is true that showing there are no solutions to $x^3+y^3=z^3$ is the easiest case of Fermat's last theorem, but it still seems harder than exhaustion here. Both will work. The second is more useful going forward.

Ross Millikan
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The only cubic residues mod 7 are 0 and $\pm 1$. So if $a^3+b^3=c^3$, either

  1. $c=7$ and $a=3,5$,or $6$ (the positive integers less than 7 whose cubes are -1 mod 7).
  2. $a=7$.

That's 3+9=12 cases to check by exhaustion, which is not so bad.

user7530
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