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I do believe that it has something to do with Fermat's little theorem, and I can prove the backward relationship, but how to prove the forward if? like is N^k-1 necessarily be n ^ 4m?

巴阿宝
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    It's not true. For example, try $n=10,;k=2$. – quasi Nov 06 '17 at 01:54
  • Or simply $n=1$ and $k$ is anything. – Isaac Browne Nov 06 '17 at 01:55
  • I did not interpret the the question to be ($n^k \equiv n \mod 5 \iff k \equiv 1 \mod 4$) for all integer $n$ and $k$ natural which obviously is not true. But as for integer $n$ and natural $k$ ($n^k \equiv n\mod 5$ for all $n \iff k \equiv 1 \mod 4$) which is true. $10^2 \equiv 10 \mod 5$ but then if we try any other $n$, $n^2 \not \equiv n \mod 5$ in general. And $1^k \equiv 1 \mod 5$ for all $k$ but that certainly not the case for any $n$ that is not a multiple of $5$. – fleablood Nov 06 '17 at 16:22

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Just try $n =2$ if $k \not \equiv 1 \mod 4$ then $2^k \not \equiv 2 \mod 5$

$\implies$

Suppose $n^k \equiv n\mod 5$ for all $n$. THen $2^k \equiv 2\mod 5$. By FLT we know $2^4 \equiv 1 \mod 5$ and if $k \equiv i \mod 4$ then $2^k\equiv 2^i \mod 4$.

We can simply try $2^0, 2^2, 2^3 \equiv 1, 4, 3 \mod 5$ and not $2$. So $i \equiv k \equiv 1 \mod 4$

$\Leftarrow$

If $\gcd(n, 5) = 1$ then by FLT then for $k \equiv 1 \mod 4$, $n^{4} \equiv 1 \mod 5$ and thus $n^{1 + 4m} \equiv n \mod 5$.

If $\gcd(n, 5) \ne 1$ then $5|n$ and $n\equiv 0 \equiv n^k \mod 5$.

fleablood
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