How do I prove that there is infinite set of numbers $m$ such that the biggest prime divisor of $m^4+1$ is bigger than $2m$?
There are infinitely many $m$ such that $m^4 + 1$ has large prime factors
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number-theory
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0In$ $that case, your best bet is this site: http://www.artofproblemsolving.com/Forum/viewforum.php?f=87 and the books on clever tricks listed there. From what I can see, the main shortcoming is that they refuse to talk about quadratic residues or quadratic reciprocity, so sometimes the answers are more wordy than we would want. Anyway, the site is large, there are number theory questions in many places. – 2012-07-18
1 Answers
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Let $p$ be a prime congruent to 1 modulo 8 --- there are infinitely many such. For each such prime, there are 8 numbers $a$, $0\lt a\lt p$, such that $p$ divides $a^8-1=(a^4-1)(a^4+1)$. So there are four numbers $m$ such that $p$ divides $m^4+1$. They come in pairs that add up to $p$, so two of them are less than $p/2$. Either one is thus an $m$ with a prime divisor exceeding $2m$.
Each $m$ has only finitely many $p$ dividing $m^4+1$, so there are infinitely many such $m$.
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0Very nice! ${}{}{}$ – 2012-07-19