# There are infinitely many $m$ such that $m^4 + 1$ has large prime factors

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 $2m$?

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Clever problem. Where did you find it? – Will Jagy Jul 18 '12 at 21:50
Someone gave it to me – Frank Jul 18 '12 at 21:58
In that case, do you know that the thing to be proved is definitely true? It seems quite likely, however...Note that we do not know for a fact that there are infinitely many primes of the form $n^2 + 1,$ although everyone believes it must be true. My point is that it is quite easy to write down plausible number theory conjectures with no hope of any proof. The other extreme is that your friend knows exactly where this came from, in some book where a successful method of attack is discussed in general terms. – Will Jagy Jul 18 '12 at 22:04
@WillJagy On the other hand, Iwaniec showed that $n^2+1$ is infinitely often $P_2$, and therefore has a prime factor larger than $n$. This question seems to be at least within reach of those techniques, although one hopes for a less technically demanding solution. – Erick Wong Jul 18 '12 at 22:16
I have never seen this before. Your choice of language suggested to me that the person who gave this to you already knew the answer. I do not. However, I know that all prime factors $p$ of any $n^2 + 1$ satisfy $p \equiv 1 \pmod 4.$ I also know that it is very difficult to have all prime factors of a number very small. So, again, you should make an effort to find out where this problem comes from originally. It may be elementary, it may be within current methods but difficult, it may be open. You have not given us enough information. – Will Jagy Jul 18 '12 at 22:29

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 $m$.
Very nice! ${}{}{}$ – Jyrki Lahtonen Jul 19 '12 at 10:32