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Prove that if a, m, n are positive integers with $m ≠ n$, then $gcd({a^2}^n+1,{a^2}^m+1)=1$ if $a$ is even and $2$ if a is odd. Use this to show that there are infinitely many primes.

WHAT?! i do not understand how this result can possibly help me prove the existence of infinite primes. I do not get how i can work out the gcd of the numbers, let alone the second part. I am extremely confused. Please help. Would be amazing if you could utilise standard number theory and/or modular arithmetic.

Aayam Mathur
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  • This probably involves Fermat's little theorem for the gcd part. For the other part, once you know the facts about gcd, consider what primes can divide the infinite sequence ${a^{2^i}+1}$. – rogerl Jun 11 '20 at 18:17
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    Just logically, if we have an infinite sequence $a_n>1$ with $\gcd(a_n,a_m)=1$ if $n\neq m$ then there have to be infinitely many primes. After all, if there were only $N$ primes then the numbers $a_1, a_2,\cdots, a_{N+1}$ couldn't all be pairwise relatively prime – lulu Jun 11 '20 at 18:17
  • Duplicate of this – Bill Dubuque Jun 10 '24 at 14:37

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  1. If $m<n$ and $d\in \Bbb N$ and $d$ divides both $a^{2^m}+1$ and $a^{2^n}+1$ then modulo $d$ we have $$0\equiv a^{2^n}+1 \equiv (a^{2^m})^{2^{n-m}}+1\equiv(-1)^{2^{n-m}}+1 \equiv 2.$$ So $0\equiv 2 \mod d,$ so $d=1$ or $d=2.$ But $d$ is odd because it is a divisor of $a^{2^m}+1,$ which is odd because $a$ is even.

  2. Any natural number greater than $1$ is divisible by a prime. For each $m\in \Bbb N$ let $d_m$ be the least (or any) prime divisor of $a^{2^m}+1.$ By 1., if $m<n$ then $d_m\ne d_n.$

DanielWainfleet
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