Square-free value of $n^2+1$ without large prime factors It is well-known that there are infinitely many square-free numbers of the form $n^2+1,n\in\mathbb{Z}$.
Question: Are there infinitely many square-free numbers of the form $n^2+1$, each with all its prime factors $\leq n$?
 A: Yes, there are infinitely many such $n$.
One way to prove this is to start from a quadratic polynomial $N(x)$ 
with integer coefficients such that there's a factorization 
$N^2+1=AB$ into two quadratics; let's use
$$
N(x) = 5x^2+x-1,
\quad
N(x)^2 + 1 = (5x^2-4x+1) (5x^2+6x+2) = A(x) B(x).
$$
Now substitute suitable positive integers for $x$, and try $n=N(x)$.
One of $A,B$ is already smaller than $N$, 
and the other can be made to have all its prime factors smaller than $N$ 
by choosing $x$ in some congruence class; 
in our case $A<N$, and $B$ works as long as $x$ is even $-$ 
which fortunately is still consistent with $N^2+1$ being squarefree in our case.
(NB $n^2+1$ could never be divisible by $4$ however $n$ was chosen.)
I claim that as $X \to \infty$ a positive fraction of even choices of $x<2X$
make $n^2+1$ squarefree.  Indeed let $S_p$ be the number of even $x < 2X$
for which $p^2 | N(x)^2 + 1$.  Then $S_p=0$ if $p>10X$ or $p \neq 1 \bmod 4$,
and otherwise $S_p < 4X/p^2 + 4$ (there are $4$ congruence classes for
$x \bmod p^2$ that must be excluded).  The sum of $4X/p^2$ over all
$p \equiv 1 \bmod 4$ converges to $cX$ for some $c<1$ (in fact $c < 1/4$
if I computed right) and the sum of $4$ over $p<10X$ is $O(X/\log X)$.
Hence $\sum_p S_p = (c+o(1)) X$ for large $X$, and we conclude that
the number of choices of $x$ that make $n^2+1$ squarefree is at least
$(1-c-o(1))X \to \infty$, QED.
