fermat numbers

Proofs from THE BOOK by Martin Aigner and Günter Ziegler begins by giving six proofs of the infinity of primes. We will go over the third proof. Before we go over this proof, lets cover some background.

— 1. Fermat Numbers —

Fermat numbers are defined by

$\displaystyle F_{n}=2^{2^{n}}+1,$

so that ${F_{0}=3,F_{1}=5,F_{2}=17,F_{3}=257,}$ and ${F_{4}=65537}$ . They are of great interest in many ways: for example, it was proved by Gauss that, if ${F_{n}}$ , is a prime ${p}$ , then a regular polygon of ${p}$ sides can be inscribed in a circle by Euclidean methods. The property of the Fermat numbers which is relevant here is

Theorem 1 No two Fermat numbers have a common divisor greater than 1.

We will prove this theorem later.

The first four Fermat numbers are prime, and Fermat conjectured that all were prime. Euler, however, found in 1732 that

$\displaystyle F_{5}=2^{2^{5}}+1=641\cdot6700417$

is composite.

In 1880 Landry proved that

$\displaystyle F_{6}=2^{2^{6}}+1=274177\cdot67280421310721.$

It is currently known that ${F_{n}}$ , is composite for ${5 \leq n \leq 32}$ . Factoring Fermat numbers is extremely difficult as a result of their large size. ${F_{12}}$ has ${5}$ known factors with ${C1187}$ remaining (where ${C}$ denotes a composite number with $n$ digits). ${F_{13}}$ has ${4}$ known factors with ${C2391}$ remaining. ${F_{14}}$ has no known factors but is composite. There are currently four Fermat numbers that are known to be composite, but for which no single factor is known: ${F_{14},F_{20},F_{22},}$ and ${F_{14}}$ . In all the other cases proved to be composite a factor is known. No prime ${F_{n}}$ has been found beyond ${F_{4}}$ , and it seems unlikely that any more will be found using current computational methods and hardware.

— 2. Infinitude of Primes Theorem —

We are now ready to prove Euclid’s Second Theorem, also called the Infinitude of Primes Theorem using the third proof in Proofs from THE BOOK.

Theorem 2 (Euclid’s Second Theorem) The number of primes is infinite.

Proof: Next let us look at the Fermat numbers ${F_{n}=2^{2^{n}}+1}$ for ${n=0,1,2,\cdots}$ . We will show that any two Fermat numbers are relatively prime (Theorem 1); hence there must be infinitely many primes. To this end, we verify the recursion

$\displaystyle \prod_{k=0}^{n-1}F_{k}=F_{n}-2\quad(n\geq1),$

from which our assertion follows immediately. Indeed, if ${m}$ is a divisor of, say, ${F_{k}}$ and ${F_{n}}$ ${(k<n)}$ , then ${m}$ divides ${2}$ , and hence ${m=1}$ or ${2}$ . But ${m=2}$ is impossible since all Fermat numbers are odd.

To prove the recursion we use induction on ${n}$ . For ${n=1}$ we have ${F_{0}=3}$ and ${F_{1}-2=5-2=3}$ . With induction we now conclude

$\displaystyle \begin{array}{rcl} \prod_{k=0}^{n} F_{k} & = & \left(\prod_{k=0}^{n-1}F_{k}\right)F_{n}\\ & = & (F_{n}-2)F_{n}\\ & = & (2^{2^{n}}-1)(2^{2^{n}}+1)\\ & = & 2^{2^{n+1}}-1\\ & = & F_{n+1}-2. \Box \end{array}$

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