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CMC - Problem 6

Problem 6. (7 points) Find the smallest prime factor of \(4294967297\).

Note by Cody Johnson
3 years, 11 months ago

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3 votes

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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.

Patrick Hompe - 3 years, 11 months ago

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+1 for the first sentence.

Ahaan Rungta - 3 years, 11 months ago

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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 \(k\cdot2^{n+2}+1,\text{ }k\in\mathbb{N}\)?

Cody Johnson - 3 years, 11 months ago

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641

Ahaan Rungta - 3 years, 11 months ago

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

Ahaan Rungta - 3 years, 11 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, 5 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, 11 months ago

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

Cody Johnson - 3 years, 11 months ago

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the 1 point or 7 points?

Michael Tang - 3 years, 11 months ago

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@Michael Tang Look at the updated rules.

Cody Johnson - 3 years, 11 months ago

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"Brilliant solution"

Lol pun

Daniel Wang - 3 years, 11 months ago

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