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Find all positive integers $n \le 6$, such that the equation $a^n+b^n=c^n+n$ has solutions over the integers.

My attempt:

For $n=1$, trivially there are integer solutions.

For $n=2$, we have the solutions $$a=b=\pm 1, c=0.$$

For $n=3$, we have the solution $$a=b=1, c = -1.$$

Anyone can complete the proof for $n=4,5,6$?

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  • $\begingroup$ You have a (small, closable) gap for $n=4,6$ because youonly supposed that $|c|$ is not the smallest. $\endgroup$ Jun 4, 2017 at 20:27
  • $\begingroup$ Yes, almost forgot. Will fix it! Thanks! $\endgroup$
    – velut luna
    Jun 4, 2017 at 20:30
  • $\begingroup$ The gap seems not so small to me--you are actually supposing $|c|$ is not the largest, rather than that $|c|$ is not the smallest. And the case that $|c|$ is the largest looks quite nontrivial. $\endgroup$ Jun 4, 2017 at 20:31
  • $\begingroup$ @EricWofsey Yes, you are right. Thanks! Edited my question. $\endgroup$
    – velut luna
    Jun 4, 2017 at 20:46
  • $\begingroup$ You have answered your question yourself here, and then deleted it. Why? $\endgroup$ Jun 5, 2017 at 19:45

2 Answers 2

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$n=4$: fourth powers are $\equiv 0$ or $1\pmod{8}$. There is no way to have $(0\text{ or }1)+(0\text{ or }1)\equiv (0\text{ or }1)+4\pmod 8$.

$n=5$: Fifth powers are $\equiv 0$ or $1$ or $-1\pmod{11}$. Three such numbers combined cannot produce $5$, i.e., $a^5+b^5-c^5\equiv -3,-2,-1,0,1,2,3\not\equiv 5\pmod {11}$.

$n=6$: Sixth powers are $\equiv 0$ or $1$ or $-1\pmod{13}$. Three such numbers combined cannot produce $6$, i.e., $a^6+b^6-c^6\equiv -3,-2,-1,0,1,2,3\not\equiv 6\pmod {13}$.

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If $n$ is such that $2n+1$ is prime, then the only $n$th powers mod $2n+1$ are $0$ and $\pm 1$ (since the group of units mod $2n+1$ is cyclic of order $2n$). It follows easily that $a^n+b^n=c^n+n$ has no solutions mod $2n+1$ if $n>3$. In particular, this handles the cases $n=5$ and $n=6$ (as well as infinitely many other larger values of $n$).

For $n=4$, you need a more ad hoc argument, since $2n+1$ is not prime. Working mod $8$ does the trick, since the only $4$th powers (in fact, the only squares) mod $8$ are $0$ and $1$.

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  • $\begingroup$ Thank you for the answer. But I need to learn more about group theory first to fully understand it! :) $\endgroup$
    – velut luna
    Jun 4, 2017 at 21:01

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