# Fermat Numbers as a product

We are discussing Fermat numbers in class, and one of the claims brought up is as follows:

"For any integer $n \ge 1$, the $n$th Fermat number is $F(n)$ = $2 + \prod_{i=0}^{n-1}F(i)$."

I have not been able to find any proofs online, but I would really like to know how a Fermat number can be represented in this form (as 2 plus the product of past Fermat numbers). Are Fermat numbers not prime, or have I confused myself? It doesn't seem intuitive to me, but I could be wrong.

• Hint: what's $(x-1)(x+1)$? How about $(x-1)(x+1)(x^2+1)$? It's also not true that Fermat numbers are prime. The small ones up to $F(4)$ are prime, but after that not a single more Fermat prime has been found (and very few are expected). Finally, what do you find about this equation that contradicts the possibility that they are prime? Jan 30, 2013 at 4:17

$$(x^{2^n}-1)=(x^{2^{n-1}}+1)(x^{2^{n-1}}-1)=(x^{2^{n-1}}+1)(x^{2^{n-2}}+1)(x^{2^{n-3}}-1)$$ $$=(x^{2^{n-1}}+1)(x^{2^{n-2}}+1)(x^{2^{n-3}}+1)(x^{2^{n-4}}+1)...(x^4+1)(x^2+1)(x+1)(x-1)$$ $$\frac{x^{2^n}-1}{x-1}=\prod_{k=0}^{n-1}(x^{2^k}+1)$$ $$2^{2^n}-1=\prod_{k=0}^{n-1}(2^{2^k}+1)$$ $$2^{2^n}+1=2+\prod_{k=0}^{n-1}(2^{2^k}+1)$$ $$F_n=2+\prod_{k=0}^{n-1}F_k$$
Let $f_n = 2^{2^n}+1$.
There is an obvious recurrence relation: $f_{n+1} = (f_n-1)^2 + 1$.
Because $f_0$ is odd and greater than $1$, any prime dividing $f_n$ cannot divide $f_{n+1}$, meaning that the Fermat numbers are all relatively prime, although not necessarily primes themselves.