Exercise 846Number theory › CongruencesL1Added 3 Nov 2026

Generalization of Fermat

#Fermat#generalization#coprime

Statement

Show that for any prime $p$ and any integer $a$ (including $p\mid a$ ), $a^p\equiv a\pmod p$ .

Hint

If $p\nmid a$ , multiply Fermat by $a$ . If $p\mid a$ , both sides vanish.

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Case $p\nmid a$ . Fermat gives $a^{p-1}\equiv1\pmod p$ . Multiply by $a$ : $a^p\equiv a\pmod p$ .

Case $p\mid a$ . $a\equiv0\pmod p$ , so $a^p\equiv0\equiv a\pmod p$ .

Conclusion. $a^p\equiv a\pmod p$ for all $a$ .

Reformulation. For prime $p$ , $X^p-X\equiv0\pmod p$ for all $X$ ; equivalently $X^p-X=X(X-1)\cdots(X-p+1)$ mod $p$ .