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I already solved the question myself, but there is something bottering me. In the exersice is told "Solve, by easy estimations". I couldnt find a boundary for $n$ or something. I started with $a^n = 8 + b^n$ and then follows $a \leq \sqrt[n]{8} + b$, so $a-b \leq \sqrt[n]{8}$. Knowing that $a$ and $b$ in $\mathbb{Z}$ we find that for $n > 3 \implies a-b \leq 1$. But now $a$ and $b$ can be negative aswell. So it isn't a boundary yet. Hopefully help comes soon, otherwise i cant sleep..

Zhanxiong
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S. Ferwerda
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  • If $n$ is even, negativeness doesn't matter. If $n$ is odd, and $a$ and $b$ are of different signs, that makes the absolute value of the difference big. – André Nicolas Dec 07 '15 at 19:18
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    If $n\ge 4$ then every one-step difference of $x^n$ except from $-1$ to $0$ and $0$ to $1$ is at least $2^4-1^4=15$, which is too large. So only $n\in{2,3}$ need to be investigated. – hmakholm left over Monica Dec 07 '15 at 19:22
  • Thanks Henning, so the boundary of $n$ is found by checking the one-step difference. Is 2^n - 1^n always the lowest one-step difference? If so i totally get it, thanks! – S. Ferwerda Dec 07 '15 at 19:26
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    just a little tidbit, adding on to Henning's answer: for the case n=3, we can see that $2^3 + b^3 = a^3$ will not have any solutions due to fermat's last theorem – pMarkov Dec 07 '15 at 19:29
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    Smart use of fermat. Only thing left is n = 2, which is pretty easy and gives all solutions. Thanks guys! – S. Ferwerda Dec 07 '15 at 19:50

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