Prove $n^7$ - n is divisible by 42 for all n.

Question

This question is pulled from a number theory practice set that did not provide an answer key. Link: https://www.math.toronto.edu/~herzig/putnam_nt_oct07.pdf

My reasoning is to prove that $n^7$ - n is divisible by 2,3, and 7 because they are the prime factors of 42. I've proven divisibility by 2, but I cannot prove divisibility by 3 or 7.

Please let me know if this is the right path, and any ideas are welcomed proving divisibility by 3 or 7.

$ n^7-n=n(n^6-1)=n(n^3+1)(n^3-1)=n(n+1)(n^2-n+1)(n-1)(n^2+n+1)$ — hamam_Abdallah, Jun 17 '21 at 22:17
Hints: By Fermat's little theorem, $n^2 \equiv 1 \bmod 3$ and $n^7 \equiv n \bmod 7$ — Keith Backman, Jun 17 '21 at 22:21
@KeithBackman. Only if $\gcd(n,3)=1$. BUt in general. Either $p|n$ (and so $p|n^k-n$) or $n^{p-1} \equiv 1\pmod p$. And if $n^{p-1}\equiv 1$ then $n^{k{p-1}} \equiv 1\pmod p$ and $n^{k{p-1} + 1} \equiv n \pmod p$. And for $p = 2,3,7$ wehave $6(2-1)+ 1=3(3-1) + 1 = 1(7-1) + 1 = 7$ so we have $n^7-n \equiv 0 \pmod{2,3,7}$ — fleablood, Jun 17 '21 at 23:00

score 0 · Accepted Answer · answered Jun 17 '21 at 22:39

0

Hint:

Use Euler-Fermat:

For any prime $p$, one has $\;n^p\equiv n\mod p$.

answered Jun 17 '21 at 22:39

Bernard

179,256

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Prove $n^7$ - n is divisible by 42 for all n.

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