= Solution
First prove $a^p\equiv a\pmod p$ for every <nonnegative integer> $a$ by <mathematical induction>. The case $a=0$ is immediate. If it holds at $a$, the <binomial theorem> and the stated divisibility of the middle <binomial coefficients> give
$$
(a+1)^p\equiv a^p+1\equiv a+1\pmod p.
$$
This establishes the induction. For an arbitrary <integer> $a$, take a nonnegative representative $r$ of its <residue class>; then $a^p\equiv r^p\equiv r\equiv a\pmod p$. When $a\not\equiv0\pmod p$, part (b) supplies a <multiplicative inverse> modulo $p$. Multiplying the congruence by that inverse proves <Fermat's little theorem>:
$$
\boxed{a^{p-1}\equiv1\pmod p.}
$$
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