Is my proof below original?
We prove Fermat's little theorem in the reduced formulation: If $p$ is an odd prime and $a$ is an integer such that $p < a$ and $p \nmid a,$ then $a^{p-1} \equiv 1\ (mod\ p).$
Note, since $p$ is an odd prime, that the congruence $$a^{p-1} \equiv 1\ (mod\ p)$$ is equivalent to the congruence $$(a^{\frac{p-1}{2}} - 1)(a^{\frac{p-1}{2}} + 1) \equiv 0\ \ (mod\ p).$$ If $$p \nmid (a^{\frac{p-1}{2}} - 1)\ \ \ and\ \ \ p \nmid (a^{\frac{p-1}{2}} + 1),$$ then there exist integers $q, r, q', r' > 0$ such that \begin{eqnarray*} a^{\frac{p-1}{2}} &=& pq + r + 1,\ \ 1 \leq r \leq p-1,\\ a^{\frac{p-1}{2}} &=& pq' + r' - 1,\ \ 1 \leq r' \leq p-1, \end{eqnarray*} so that $$p(q' - q) = r - r' + 2,$$ whence $$p \mid (r - r' + 2),$$ and we have $$r - r' \geq p - 2.$$ So, since $1 \leq r, r' \leq p-1,$ certainly $$r - r' = p-2.$$ With no loss of generality, put $$r := p-1,\ \ r' := 1.$$ But then $$p \mid a^{\frac{p-1}{2}};$$ a contradiction.
