How to solve these two simultaneous "divisibilities" : $n+1\mid m^2+1$ and $m+1\mid n^2+1$

Solution 1:

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.

Solution 2:

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.

$$m=2x-1$$

$$n=2y-1$$

$$z=9x-2y+2$$

$$k=9y-2x+2$$