Greatest prime divisor of $ n^2+1$ Prove that there exist infinitely many positive integers $n$ such that the greatest prime divisor of $n^2+1$ is less than $n \cdot \pi^{-2019}.$
What did I do: $n = 2a^2$, then $n^2+1 = 4a^4+1 = ((a-1)^2+a^2)((a+1)^2+a^2)$. 
 Funny that it's trivial for $P (n ^ 2 + 1) <n$, because all n satisfy. But $\pi^{-2019}.$ goes down a lot .... how do I prove it?
 A: Let us prove that for every $\varepsilon > 0$ there exist infinitely many integers $n > 0$ such that $P(n^2 + 1) < \varepsilon n$. 
Obviously, it is enough to prove that there exists one such $n$. As you noticed, if $n = 2m^2$, for some integer $m > 0$, then $n^2 + 1 = (2m^2 - 2m + 1)(2m^2 + 2m + 1)$. Hence, it is enough to prove that we can find $m$ such that $P(2m^2 - 2m + 1) < 2\varepsilon m^2$ and $P(2m^2 + 2m + 1) < 2\varepsilon m^2$. 
It is well known and not difficult to prove that for every polynomial $f \in \mathbb{Z}[x]$ the set of prime numbers $p$ such that $p$ divides $f(k)$ for some integer $k > 0$ is infinite. Therefore, we can find an integer $\ell > 0$ such that $p_1 := P(2\ell^2 - 2\ell + 1) > 2/\varepsilon$ and $p_2 := P(2\ell^2 + 2\ell + 1) > 2/\varepsilon$. 
Now letting $m := \ell + t p_1 p_2$, for some integer $t > 0$, we get that $p_1$ divides $2m^2 - 2m + 1$ and $p_2$ divides $2m^2 + 2m + 1$. 
Hence, $P(2m^2 - 2m + 1) < \max(p_1, (2m^2 - 2m + 1) / p_1)$ and $P(2m^2 + 2m + 1) < \max(p_2, (2m^2 + 2m + 1) / p_2)$. 
Taking $t$ sufficiently large, we get that $P(2m^2 + 2m + 1) < (2m^2 + 2m + 1) / p_2 < \varepsilon(2m^2 + 2m + 1) / 2 < \varepsilon 4m^2 / 2 < 2\varepsilon m^2$ and, similarly, $P(2m^2 - 2m + 1) < 2\varepsilon m^2$, as required.
