# A possible proof of Fermat's Little Theorem

By the formula $$(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}$$ we know that $(a+b)^p\equiv a^p+b^p \pmod{p}$,

Is there a proof of Fermat's Little Theorem based on this fact?

• Yes, by induction. – Lord Shark the Unknown Jul 5 '18 at 20:13
• My lord I'll try!!!!!!!!!! – Leaning Jul 5 '18 at 20:13
• @LordSharktheUnknown Induction? Not the fact that $p$ divides $p$ choose $k$? – Sorfosh Jul 5 '18 at 20:22
• @Sorfosh: The OP already has established $(a+b)^p\equiv a^p+b^p \pmod p$ using the fact you mentioned. The issue at hand is how to derive Fermat's Little theorem. That will, in fact, be a proof by induction — which appears in many, many algebra texts. – Ted Shifrin Jul 5 '18 at 20:30
• See also: Deriving Fermat's little theorem from $(a+1)^p\equiv a^p+1$ modulo $p$? (The question is at least similar, maybe it could be considered even a duplicate.) – Martin Sleziak Jul 6 '18 at 7:29

As noted in the comment by Lord Shark, using that fact, for the induction step, assuming as hypotesis $a^p\equiv a \mod p$ we have
$$(a+1)^p\equiv a^p+1\equiv a+1 \mod p$$