Solution
= Solution
\b[False.] The <Carmichael number> $561=3\cdot11\cdot17$ supplies a counterexample. For every $a$ coprime to $561$, the <Fermat little theorem> gives $a^{560}\equiv1$ modulo each of these primes, since $2,10,16$ all divide $560$. The <Chinese remainder theorem> then gives $a^{560}\equiv1\pmod{561}$. Thus the proposed <Fermat primality test> can fail to detect a composite for every coprime base.