How to solve these two simultaneous "divisibilities" : $n+1\mid m^2+1$ and $m+1\mid n^2+1$ Is it possible to find all integers $m>0$ and $n>0$ such that $n+1\mid m^2+1$ and $m+1\,|\,n^2+1$ ?
I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.
Thanks !
 A: Some further results along the lines of thought of @individ:
Suppose $p$ and $s$ are solutions to the Pell's equation:
$$-d\cdot p^2+s^2=1$$
Then,
\begin{align}
m &= a\cdot p^2+b\cdot pq +c\cdot q^2\\
n &= a\cdot p^2-b\cdot pq +c\cdot q^2
\end{align}
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
\begin{align}
(10,4,-2,-15)\\
(39,12,-3,-65)\\
\end{align}
Sadly, the solutions are negative.
Here are some examples:
\begin{align}
(m,n) &= (-6,-38)  &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\\
(m,n) &= (-290,-2274)  &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\\
(m,n) &= (-15171,-64707)  &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\\
(m,n) &= (-1009692291,-4306907523)  &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\\
(m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\\
\end{align}
P.S. I am also very curious how @individ thought of this parametrization. 
A: You can record a similar system:
$$\left\{\begin{aligned}&m^2+t^2=(n+t)z\\&n^2+t^2=(m+t)k\end{aligned}\right.$$
Parametrization of solutions we write this.
$$m=q(3x-q)$$
$$n=2x^2-qx-q^2$$
$$t=3q^2-3xq+2x^2$$
$$z=5q^2-2qx+x^2$$
$$k=5q^2-8qx+4x^2$$
Consider a special case.
$$\left\{\begin{aligned}&m^2+1=(n+1)z\\&n^2+1=(m+1)k\end{aligned}\right.$$
Using the solutions of the equation Pell.
$$p^2-15s^2=1$$
Enough to know first, everything else will find a formula.  $(p;s) - (4;1)$
$$p_2=4p+15s$$
$$s_2=p+4s$$
The solution then write.
$$m=-2p^2-4ps+10s^2$$
$$n=-2p^2+4ps+10s^2$$
$$z=8m+9-n$$
$$k=8n+9-m$$
These solutions are negative.
And a positive decision of the same are determined by the Pell equation. 
$$p^2-65s^2=-1$$
Use the first solution.  $(p;s) - (8;1)$
Next find the formula.
$$p_2=129p+1040s$$
$$s_2=16p+129s$$
Will make a replacement.
$$x=p^2+6ps+13s^2$$
$$y=p^2-6ps+13s^2$$
The decision record.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
A: Trying again, now as answer instead of just a comment:
I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?
The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.
simultaneouos_divisions_MSE.pdf
A: To solve this system of equations - it is necessary to solve the system.
$$\left\{\begin{aligned}&m^2+t^2=(n+t)z\\&n^2+t^2=(m+t)k\end{aligned}\right.$$
It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.
$$m=33643p^2\pm5404ps+217s^2$$
$$n=5491p^2\pm852ps+33s^2$$
$$t=s^2-153p^2$$
$$z=212041p^2\pm34274ps+1385s^2$$
$$k=901p^2\pm134ps+5s^2$$
We need a case of when.  $t=s^2-153p^2=1$
Knowing the first decision  $(s ; p ) - (2177;176)$
The rest can be found by the formula.
$$s=2177s+26928p$$
$$p=176s+2177p$$
Although this equation can be not enough. We need to find when there are multiple solutions.
