# CMC - Problem 6

Problem 6. (7 points) Find the smallest prime factor of $4294967297$. Note by Cody Johnson
6 years, 7 months ago

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This is pretty silly. This is the 5th fermat number, which is $2^{32}+1$. It can be written as $2^{28}(5^4+2^4)-(5 \cdot 2^7)^4+1=2^{28} \cdot 641 -(640^4-1)$, so it's clear that 641 divides it. For another prime that divides it, we must have $2^{32} \equiv -1 \pmod{p}$, and therefore $2^{64} \equiv 1 \pmod{p}$, so by Fermat's Little Theorem we must have that 64 divides p-1, or that p=64k+1. Now there are 9 possibilities, but some of them aren't prime. It's easy to check them.

- 6 years, 7 months ago

+1 for the first sentence.

- 6 years, 7 months ago

6 points for Patrick Hompe! -7 points for calling it silly! >:O Did you know that in general, each prime factor of $F_{n\ge2}$ is in the form of $k\cdot2^{n+2}+1,\text{ }k\in\mathbb{N}$?

- 6 years, 7 months ago

641

- 6 years, 7 months ago

Python:

def primes(n):
if n<=2:
return []
sieve=[True]*(n+1)
for x in range(3,int(n**0.5)+1,2):
for y in range(3,(n//x)+1,2):
sieve[(x*y)]=False

return +[i for i in range(3,n,2) if sieve[i]]


- 6 years, 7 months ago

Euler showed that this is the first composite Fermat number :D 641 divides it

Also, all the prime factors of $a^{2^n}+b^{2^n}$ are of the sort $1+k.2^{n+1}$

- 6 years, 1 month ago

Why does this have the "points exchange" tag? Also, is this possible to do without W|A? If somebody finds it without W|A, it's pure luck. Also, fun fact: my FTC Robotics team ID number is 7297.

- 6 years, 7 months ago

Yes, there's a brilliant solution to this. The answer is 641, so you get the 1 point.

- 6 years, 7 months ago

the 1 point or 7 points?

- 6 years, 7 months ago

Look at the updated rules.

- 6 years, 7 months ago

"Brilliant solution"

Lol pun

- 6 years, 7 months ago