Prove that $7\mid n^7+7n^3-n$ for every positive integer n.

Question

Let $P(n)=7\mid n^7+7n^3-n$ for integers $n\ge 1$. Then $P(1)$ is true because $1+7-1=7$ is multiple of $7$.

Assume that $P(n)$ is true and $n\in \mathbb{Z}$ and $n\ge 1$. Then $P(n+1)$ is true because;

$n^7+7n^3-n=7m$ for some natural number $m$

So, $$(n+1)^7+7(n+1)^3-(n+1)=...$$

I'm stuck here, and I can't continue the proof.

Can you help?

@ucei The title doesn't say that; there is more to the right of the $|$ bar. — coffeemath, Dec 18 '20 at 22:22

score 5 · Answer 1 · answered Dec 18 '20 at 22:27

5

HINT: Instead of induction here, we can use Fermat's Little Theorem as it says $n^7 \equiv n \mod 7$ for every positive integer $n$.

answered Dec 18 '20 at 22:27

ArsenBerk

13,388

score 2 · Accepted Answer · answered Dec 18 '20 at 22:28

Note that $7 \vert 7n^3$ so the problem reduces to proving $n^7 \equiv n \pmod 7$. This is obviously true for $n=1$.

Assume $k^7 \equiv k \pmod 7$. Then $(k+1)^7 = \sum_{i=0}^7 \binom 7i k^i \equiv k^7+1 \pmod 7$ because $7 \vert \binom 7i$ for $ 1 \leq i \leq 6$. By our inductive hypothesis, $k^7 \equiv k \pmod 7$, so $(k+1)^7 \equiv k^7+1 \equiv k+1 \pmod 7$ and we are done.

Prove that $7\mid n^7+7n^3-n$ for every positive integer n.

2 Answers2