TITLE: Solve in integers: $y(x^2+1)=z^2+1$
QUESTION [9 upvotes]: Find all integer solutions to the equation
$$
y(x^2+1)=z^2+1.
$$
There is, for example, an infinite family of solutions $x=u$, $y=(uv\pm1)^2+v^2$, $z=(u^2+1)v \pm u$, $u,v \in {\mathbb Z}$, but there are also solutions outside of this family, e.g. $(x,y,z)=(8,5,18)$ or $(x,y,z)=(12,2,17)$. The question is to describe all integer solutions. Any reasonable description is ok. An algorithm generating all solutions is also ok, provided that it does not involve any search by trial and error (otherwise there is a trivial algorithm that tries all triples $(x,y,z)$ in some order). Using parametric expressions, recurrence relations, or something like ``start with this solution and apply these operations in any order'' (like generating Markov numbers via Markov tree) would be ideal, but more complicated algorithms are also possible. For example, for equation $yz=x^3+1$, there is an obvious algorithm "let $x$ be an  arbitrary integer, let $y$ be any divisor of $x^3+1$, and then let $z=(x^3+1)/y$", which I think is acceptable. The equation in question is one of the smallest/simplest ones for which I do not see any reasonable method/algorithm to describe all solutions, hence the question.
Remark: If we define, for any integer $u$, set $S(u)$ as a set of integers $0\leq r < u^2+1$ such that $\frac{r^2+1}{u^2+1}$ is an integer, then all the solutions to the equation are in the form $x=u$, $z=(u^2+1)v \pm r$ and $y=(u^2+1)v^2 \pm 2vr + \frac{r^2+1}{u^2+1}$ for $u,v \in {\mathbb Z}$ and $r \in S(u)$, but we need trial and error to construct set $S(u)$, so I do not think this is an acceptable answer.

REPLY [3 votes]: This answer (which I made Community Wiki) attempts an explicit summary of how to ``list" explicitly the solutions of the given equation, given the answer of Robert Israel and the comment of David Speyer which followed it.
It's convenient to deal with the prime $2$ first, but this is easy, because the only condition imposed on $z$ to get the right power of $2$ dividing $z^{2}+1$ is that if $x$ is odd, we need $z$ odd.
If $p$ is an odd prime such that $p^{n}$ is the exact power of $p$ dividing $x^{2}+1$, then we just require that $p^{n}$ divides $z^{2} -x^{2}$, so we need $ z \equiv \pm x$ (mod $p^{n}$)  since $p$ is odd.
Notice then that, in all cases, there are $2^{k}$ possible congruences for $z$ (mod $x^{2}+1$), where $k$ is the number of distinct odd primes dividing $x^{2}+1$.
To be more explicit, if $x$ is even, let the prime factorization of $x^{2}+1$ be $x^{2}+1 = \prod_{i=1}^{k}p_{i}^{n_{i}}$, where the $p_{i}$ are distinct primes, and if $x$ is odd, let the prime factorization of $\frac{x^{2}+1}{2}$ be $\frac{x^{2}+1}{2} = \prod_{i=1}^{k}p_{i}^{n_{i}}$.
For any ordered $k$-tuple of signs $(\epsilon_{1}, \ldots, \epsilon_{k})$, solve (simultaneously) $z \equiv \epsilon_{i} x$ (mod $p_{i}^{n_{i}}$) for each $i$. If $x$ is odd, also impose the condition that $z$ is odd.
In either case, this gives $2^{k}$ distinct possibilities for the congruence of $z$ (mod $x^{2}+1$) which yield $x^{2} + 1 | z^{2}+1$, and there are no other valid congruences (where, as noted earlier, $k$ is the number of distinct odd prime divisors of $x^{2}+1$).
For each integer $r$ and each integer $m \neq 0$ let $S(m,r)$ be the set of integers $0\leq w<|m|$ such that $w^2 \equiv r \,(\text{mod } m)$. We have just explicitly described the set $S(x^2+1,-1)$ for any integer $x$. Using it, the complete solution set to our equation is given by $x=u$, $z=(u^2+1)v + r$ and $y=(u^2+1)v^2 + 2vr + \frac{r^2+1}{u^2+1}$ for $u,v \in {\mathbb Z}$ and $r \in S(u^2+1,-1)$.