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.
–
Patrick Hompe
·
3 years, 8 months ago

@Patrick Hompe
–
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 \(k\cdot2^{n+2}+1,\text{ }k\in\mathbb{N}\)?
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Cody Johnson
·
3 years, 8 months ago

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

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 \(1+k.2^{n+1}\)
–
Bogdan Simeonov
·
3 years, 2 months ago

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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.
–
Ahaan Rungta
·
3 years, 8 months ago

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@Ahaan Rungta
–
Yes, there's a brilliant solution to this. The answer is 641, so you get the 1 point.
–
Cody Johnson
·
3 years, 8 months ago

## Comments

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TopNewestThis 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. – Patrick Hompe · 3 years, 8 months ago

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– Ahaan Rungta · 3 years, 8 months ago

+1 for the first sentence.Log in to reply

– Cody Johnson · 3 years, 8 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 \(k\cdot2^{n+2}+1,\text{ }k\in\mathbb{N}\)?Log in to reply

641 – Ahaan Rungta · 3 years, 8 months ago

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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 \(1+k.2^{n+1}\) – Bogdan Simeonov · 3 years, 2 months ago

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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. – Ahaan Rungta · 3 years, 8 months ago

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– Cody Johnson · 3 years, 8 months ago

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

– Michael Tang · 3 years, 8 months ago

the 1 point or 7 points?Log in to reply

– Cody Johnson · 3 years, 8 months ago

Look at the updated rules.Log in to reply

Lol pun – Daniel Wang · 3 years, 8 months ago

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