Infinitude of Primes
Via Fermat Numbers

The Fermat numbers form a sequence in the form $\displaystyle F_{n} = 2^{2^{n}} + 1,$ $n = 0, 1, 2, \ldots$ Clearly all the Fermat numbers are odd. Moreover, as we'll see shortly, any two are mutually prime. In other words, each has a prime factor not shared by any other. Hence, the number of primes cannot be finite.

That no two Fermat numbers have a non-trivial common factor follows from the two lemmas below.

Lemma 1

$F_{0}\cdot F_{1}\cdot F_{2}\cdot \ldots \cdot F_{k-1} = F_{k} - 2,$ $k \ge 1.$

Proof

The proof is by induction. Since $F_{0} = 3$ and $F_{1}=5,$ the claim indeed holds for $k = 1.$ Assume it holds for $k = n.$ For $k = n+1,$

$\begin{align} F_{0}\cdot F_{1}\cdot F_{2}\cdot \ldots \cdot F_{n}&= (F_{0}\cdot F_{1}\cdot F_{2}\cdot \ldots \cdot F_{n-1})\cdot F_{n}\\ &= (F_{n} - 2)\cdot F_{n}\\ &= (2^{2^{n}} - 1)(2^{2^{n}} + 1)\\ &= 2^{2^{n+1}} - 1\\ &= F_{n+1} - 2. \end{align}$

Lemma 2

For $n \ne m,$ $F_{n}$ and $F_{m}$ are mutually prime.

Proof

Indeed, assume $t$ divides both $F_{n}$ and $F_{m},$ $m \lt n.$ By Lemma 1,

$F_{n} = F_{0}\cdot F_{1}\cdot \ldots \cdot F_{m}\cdot \ldots \cdot F_{n-1} + 2,$

implying that $t$ divides

$F_{n} - F_{0}\cdot F_{1}\cdot \ldots \cdot F_{m}\cdot \ldots \cdot F_{n-1} = 2.$

But $2$ has only two factors: $1$ and $2.$ Being odd, Fermat numbers are not divisible by $2.$ Hence $t = 1,$ which proves the lemma.

Reference

  1. M. Aigner, G. Ziegler, Proofs from THE BOOK, Springer, 2000
  2. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth Edition, Oxford Science Publications, 1996.

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