About the congruence $x^k \equiv x \pmod 2$

Question

How can one prove that $x^k\equiv x\pmod2$ for any integer $x$ and any positive integer $k>1$.

Any integer $x$ is congruent to $0$ or $1\pmod 2.$ Can you take it from there? — J. W. Tanner, Jun 13 '21 at 07:36
In any ring $,x^2 = x\Rightarrow x^n = x,$ for all $n> 0.,$ See here for a few proofs. Here $,x^2\equiv x,$ by $,2\mid x(x-1)\ \ $ — Bill Dubuque, Jun 13 '21 at 07:46

score 2 · Accepted Answer · answered Jun 13 '21 at 07:07

2

If $x$ is even then obviously $x^k - x$ is even. If $x$ is odd then $x^k$ is odd, so $x^k - x$ even. Thus $x^k = x \mod 2$

answered Jun 13 '21 at 07:07

Infinity_hunter

1 Answers1