Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 9H / iii / Solution

False. The Carmichael number $561=3\cdot 11\cdot 17$ supplies a counterexample. For every $a$ coprime to $561$, the Fermat little theorem gives $a^{560}\equiv 1$ modulo each of these primes, since $2,10,16$ all divide $560$. The Chinese remainder theorem then gives $a^{560}\equiv 1(\mathrm{mod}561)$. Thus the proposed Fermat primality test can fail to detect a composite for every coprime base.