Infinitely many composite numbers of the form $n^2+1$, where $n$ is even? Landau's fourth problem is whether there are infinitely many primes of the form $n^2+1$. I am interested in the opposite question; are there infinitely many composite numbers of the form. If $n$ is odd, then clearly $n^2+1$ is composite. For even $n$ then. 
Link. 
 A: When $x$ ends in the digit $2$, then $x^2+1$ ends in the digit $5$.
A: By applying Hardy's technique from his "An Introduction To The Theory Of Numbers" (more details here)
$$f(n)=n^2+1 \Rightarrow f\left(\color{red}{k(n^2+1)+n}\right)=\left(k(n^2+1)+n\right)^2+1=\\
k^2\left(n^2+1\right)^2+2k\left(n^2+1\right)n+n^2+1=
\left(n^2+1\right)\left(k^2\left(n^2+1\right)+2kn+1\right), \space\forall k\in\mathbb{N}$$
Now, you can choose $n$ and $k$ to make $\color{red}{k(n^2+1)+n}$ even infinitely many times.
A: Let $n=2k^2$, with $k \geq 2$, then
$$n^2+1=4k^4+1=4k^4+1+4k^2-4k^2=(2k^2+1)^2-(2k)^2=(2k^2+1-2k)(2k^2+1+2k).$$
A: For any prime $p\equiv 1\pmod{4}$ we have that $-1$ is a quadratic residue, hence for some $a\in\left[1,\frac{p-1}{2}\right]$ we have that $x=pk\pm a$ ensures $p\mid (x^2+1)$. By invoking the Chinese remainder theorem we have that for infinite $x\in\mathbb{Z}$ the number $x^2+1$ has at least two prime factors. Actually we get that the number of prime factors of $x^2+1$, for some $x\in\mathbb{Z}$, is as large as we want.
A: We can generalize Lord Shark the Unkown's answer.  
If $k^2 + 1 = m$ then for any $a$, $(am + k)^2 + 1= a^2m^2 + 2amk + k^2 + 1 = a^2m^2 + 2amk + m = m(a^2m + 2ak + 1)$ will be a multiple of $m$.
In other words for any $x$ in the form $x = (a(k^2+1) + k)$ then $x^2 + 1$ will be divisible by $(k^2 + 1)$
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Or in other words.
If $k^2 + 1\equiv 0 \mod m$ then for any $x \equiv k \mod m$ will be have $x^2 + 1$ being divisible by $m$.
ANd there are infinite $k^2 + 1 \equiv 0 \mod m$.  Just set $m = k^2 + 1$.
