The Fermat numbers \(F_n\) are defined by \(F_n=2^{2^n}+1\), \(n\in\mathbb{Z}^+\).

Show that, for any positive integer \(k\), \(F_m\) divides \(F_{m+k}-2\), and deduce that any two distinct Fermat numbers are relatively prime.

Therefore, \(F_m\) divides \(F_{m+k}-2\).

Take any two distinct Fermat numbers, say \(F_a < F_b\). Let \(d = \gcd(F_a, F_b)\).

*The expression \(\gcd(a,b)\) stands for ‘the greatest common divisor of \(a\) and \(b\)’, or the ‘highest common factor of \(a\) and \(b\)’. *

We know \(F_a|F_b-2\) (where the vertical line means ‘divides into exactly’). Using the definition of \(d\), \(d|F_a\) and, hence, \(d|F_b-2\).

Since we also know that \(d|F_b\), it follows that \(d|F_b-(F_b-2)\), and so \(d|2\).

But all Fermat numbers are odd, and, therefore, \(d\) cannot be \(2\). So \(d=1\), and the numbers \(F_a\) and \(F_b\) are relatively prime.

Deduce that there are at least \(n+1\) distinct prime numbers less than or equal to \(F_n\), for any \(n\)…

We know that for every \(i\in \{0,1,2,...,n-1\}\), \(F_i\) and \(F_n\) are coprime.

Therefore, working upwards, each \(F_i\)’s factorisation should contain at least one new prime number smaller than \(F_n\).

Justification: Fermat numbers are increasing, and any two Fermat numbers are coprime.

So there are at least \(n\) prime numbers less than \(F_n\).

But, \(F_n\)’s factorisation should also contain a new prime, and, obviously, that prime would be less than or equal to \(F_n\).

Hence, there are at least \(n+1\) distinct prime numbers less than or equal to \(F_n\).

… and hence that there are infinitely many prime numbers.

We’ll give a proof by contradiction.

Suppose that there are finitely many prime numbers, say \(N\).

Then, if we look at \(F_N\), there are at least \(N+1\) prime numbers less than or equal to \(F_N\), which contradicts our assumption.

Hence, there are infinitely many primes.