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 !
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 !
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.
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.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
It is convenient to introduce $x:=m+1$ and $y:=n+1$. Then $$ \begin{cases} x\ |\ y^2 - 2y + 2,\\ y\ |\ x^2 - 2x + 2. \end{cases} $$ It follows that $\gcd(x,y)\mid 2$, and thus there are two cases to consider:
Case $\gcd(x,y)=1$. We have $$xy\ |\ x^2 + y^2 - 2x - 2y + 2.$$
This is solved via Vieta jumping. Namely, one can show that $\frac{x^2 + y^2 - 2x - 2y + 2}{xy}\in\{ 0, -4, 8 \}$. The value $0$ corresponds to an isolated solution $x=y=1$, while each of the other two produces an infinite series of solutions, where $x,y$ represent consecutive terms of the following linear recurrence sequences: $$1,\ -1,\ 5,\ -17,\ 65,\ \dots \qquad(s_k=-4s_{k-1}-s_{k-2}+2)$$ and $$-1, -1, -5, -37, -289,\ \dots, \qquad(s_k=8s_{k-1}-s_{k-2}+2).$$
However, there are no entirely positive solutions here.
Case $\gcd(x,y)=2$. Letting $x:=2u$ and $y:=2v$ with $\gcd(u,v)=1$, we similarly get $$ \begin{cases} u\ |\ 2v^2 - 2v + 1,\\ v\ |\ 2u^2 - 2u + 1. \end{cases} $$ Unfortunately, Vieta jumping is not applicable here. Still, if we fix $$k:=\frac{2u^2 + 2v^2 - 2u - 2v + 1}{uv},$$ then the problem reduces to the following Pell-Fermat equation: $$((k^2-16)v + 2k+8)^2 - (k^2-16)(4u - kv - 2)^2 = 8k(k+4).$$
Example. Value $k=9$ gives $$z^2 - 65t^2 = 936.$$ with $z:=65v + 26$ and $t:=4u - 9v - 2$. It has two series of integer solutions in $z,t$: $$\begin{bmatrix} z_\ell\\ t_\ell\end{bmatrix} = \begin{bmatrix} 129 & -1040\\ -16 & 129\end{bmatrix}^\ell \begin{bmatrix} z_0\\ t_0\end{bmatrix}$$ with initial values $(z_0,t_0) \in \{(39,-3),\ (-1911,237)\}$.
Not every value of $(z_\ell, t_\ell)$ corresponds to integer $u,v$. Since the corresponding matrix has determinant 260, we need to consider sequences $(z_\ell, t_\ell)$ modulo 260. It can be verified that only first sequence produces integer $u,v$ and only for odd $\ell$, that is \begin{split} \begin{bmatrix} v_s\\ u_s\end{bmatrix} &= \begin{bmatrix} 65 & 0\\ -9 & 4\end{bmatrix}^{-1}\left(\begin{bmatrix} 129 & -1040\\ -16 & 129\end{bmatrix}^{2s+1} \begin{bmatrix} 39\\ -3\end{bmatrix} + \begin{bmatrix} -26\\ 2\end{bmatrix}\right)\\ &=\begin{bmatrix} 70433 & -16512\\ 16512 & -3871\end{bmatrix}^s \begin{bmatrix} 627/5 \\ 147/5\end{bmatrix} - \begin{bmatrix} 2/5 \\ 2/5\end{bmatrix} \end{split} or in a recurrence form: \begin{cases} v_s = 70433\cdot v_{s-1} -16512\cdot u_{s-1} + 21568,\\ u_s = 16512\cdot v_{s-1} -3871\cdot u_{s-1} + 5056, \end{cases} with the initial value $(v_0,u_0) = (125, 29)$. The next couple of values is $(8346845, 1956797)$ and $(555582723389, 130248348509)$, and it can be seen that the sequence grows quite fast.
UPDATE. Series of positive solutions exist for $$k\in \{9, 13, 85, 97, 145, 153, 265, 289, 369, \dots\}.$$ Most likely, this set is infinite, but I do not know how to prove this.
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.
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.