$x^2+1$ is almost always square free It seems like $x^2+1$ is almost always square free. Any research or heuristics why?
I tried breaking the problem into solving $$x^2-ky^2=1$$
For various $k$, and I conjecture that for every $k$ there are at most finitely many solutions to the Diophantine equation. I'm pretty sure this is correct but dont know how to prove it. Any ideas on  my two problems?
 A: The equation
$$x^2-ky^2=1$$
for non-square $k$ is known as Pell's equation. Lagrange proved that every Pell's equation has infinitely many solutions. Moreover, if $(x_0,y_0)$ is the smallest non-trivial positive solution, all integer solutions are implicitly given by
$$x+y\sqrt k=\pm\,(x_0+y_0\sqrt k)^n\;, \qquad n\in\mathbb Z.$$
As you can see, the solutions generated by this formula quickly become very large, making it natural to conjecture that there are only finitely many solutions. For $k=991$, the smallest solution is
$$y=12,055,735,790,331,359,447,442,538,767,$$
which illustrates that the smallest solution can still be extremely big.
(found here, actually Joseph Rotman's A First Course in Albegra: with applications)
Note that for square $k=m^2$ the equation becomes $(x+my)(x-my)=1$, which has at most $2$ solutions.
Regarding the square-freeness of $x^2+1$, I found this paper by D.R. Heath-Brown on arXiv, giving asymptotics for the number of $x$ for which $x^2+1$ is square-free. It turns out that such numbers have non-zero asymptotic density equal to
$$\frac12\prod_{p\equiv1\pmod4}\left(1-\frac2{p^2}\right).$$
A: It is not true that $x^2+1$ is almost always square-free. For if $x\equiv 7\mod{25}$, then $x^2+1$ is divisible by $25$.
