A well-known result about Fibonacci numbers says

gcd(*F*_{m}, *F*_{n}) = *F*_{gcd(m, n)}

In words, the greatest common divisor of the *m*th and *n*th Fibonacci numbers is the *g*th Fibonacci number where *g* is the greatest common divisor of *m* and *n*. You can find a proof here.

M. Wunderlich used this identity to create a short, clever proof that there are infinitely many primes.

Suppose on the contrary that there are only *k* prime numbers, and consider the set of Fibonacci numbers with prime indices: *F*_{p1}, *F*_{p2}, … *F*_{pk}. The Fibonacci theorem above tells us that these Fibonacci numbers are pairwise relatively prime: each pair has greatest common divisor *F*_{1} = 1.

Each of the Fibonacci numbers in our list must have only one prime factor. Why? Because we have assumed there are only *k* prime numbers, and no two of our Fibonacci numbers share a prime factor. But *F*_{19} = 4181 = 37*113. We’ve reached a contradiction, and so there must be infinitely many primes.

**Source**: M. Wunderlich, Another proof of the infinite primes theorem. American Mathematical Monthly, Vol. 72, No. 3 (March 1965), p. 305.

Here are a couple more unusual proofs that there are infinitely many primes. The first uses a product of sines. The second is from Paul Erdős. It also gives a lower bound on π(*N*), the number of primes less than *N*.

The proof is wrong because F_2 = 1

Good catch! I didn’t see that, and neither did Wunderlich or his editors. So leaving out F_2, we have k primes and k-1 Fibonacci numbers bigger than 1. Finding a Fibonacci number with two prime factors doesn’t establish a contradiction. But F_37 = 24157817 = 73*149*2221 has three factors, and that does establish a contradiction.

I’m wondering: what (set of weakest) properties does a sequence need to have for the GCD-identity to be true. F = Fib works but that’s relatively arbitrary.