# 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

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

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