Exercise 2.1.20 ($n^7 -n$ is divisible by $42$)

Question

Prove that $n^7 -n$ is divisible by $42$, for any integer $n$.

Richard Ganaye · Accepted Answer

\begin{proof} Let $n$ be any integer. By Fermat's theorem, $$7 \mid n^7 -n.$$

Since $2 \mid n^2 -n$ and $3 \mid n^3 - n$ by the same theorem,
$$2 \mid n^7 - n = (n^2 - n)(n^2+n+1)(n^3+1),$$
and 
$$ 3 \mid n^7 -n = (n^3 - 1) n (n^3 +1).$$
Since $2,3,7$ are distinct primes,
$$42 = 2 \cdot 3 \cdot 7 \mid n^7-n.$$
\end{proof}

Exercise 2.1.20 ($n^7 -n$ is divisible by $42$)

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Exercise 2.1.20 ($n^7 -n$ is divisible by $42$)

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