A Fermat prime is a prime number of the form 2^{2^n}+1. The first five Fermat primes are 3, 5, 17, 257, and 65537. Fermat primes are named after Pierre de Fermat, who studied them in the 17th century.

Fermat primes are interesting because they are related to the Fibonacci sequence. The Fibonacci sequence is a sequence of numbers where each number is the sum of the previous two numbers. The first few numbers in the Fibonacci sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903.

If you take any two successive Fibonacci numbers and calculate the ratio between them, you will get a number that converges to a value called What is the largest known Fermat prime? The largest known Fermat prime is $2^{2^{64}} + 1$, which is also known as $F_{64}$. This was discovered in 2012 by the Great Internet Mersenne Prime Search (GIMPS), using a computer with an Intel Core i7-3770K CPU.

##### How do you prove a Fermat number is prime?

A Fermat number is a prime number of the form

F_n=2^{2^n}+1

The first few Fermat numbers are 3, 5, 17, 257, and 65537.

Euler proved that if n is a positive integer then

F_n equiv 2 pmod 3

So, to prove that a Fermat number is prime, it is enough to show that it is not divisible by 3.

Suppose to the contrary that F_n is divisible by 3. Then

F_n equiv 2 pmod 3 Rightarrow 2^{2^n} equiv 1 pmod 3

But this is impossible, because the order of 2 modulo 3 is 3 (i.e., 2 is a primitive root modulo 3), which means that 2 can never be congruent to 1 modulo 3.

Therefore, F_n is not divisible by 3, and hence it is prime. Are Fermat primes infinite? The answer to this question is not currently known. However, it is conjectured that there are infinitely many Fermat primes. Is 4294967297 a prime number? 4294967297 is not a prime number. It is composite, and its factors are 3 and 1430492833.

### Why are Fermat numbers coprime?

The Fermat numbers are defined as:

F_n = 2^{2^n} + 1

It can be shown that for any prime number p, if p divides F_n, then p must also divide F_m for some m < n. This is because if p is a prime number that divides F_n, then p must also divide 2^{2^n} - 1. But 2^{2^n} - 1 is a divisor of 2^{2^{n-1}} - 1, so p must also divide 2^{2^{n-1}} - 1 (because p is a prime number that divides their difference, 2^{2^n} - 1 - (2^{2^{n-1}} - 1) = 2^{2^{n-1}}). Thus, p must also divide F_{n-1}. By induction, p must divide F_m for some m < n.

Therefore, the only prime numbers that can divide any given Fermat number are those that also divide some other Fermat number. But, as it turns out, there are no such prime numbers. This is because if p were a prime number that divided F_n, then p would have to divide 2^{2^n} - 1. But 2^{2^n} - 1 is a divisor of 2^{2^{n-1}} - 1, so p would have to