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 !
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11Could you write down your proof for the infinite number of solutions? It may help everyone. – user37238 Feb 25 '16 at 8:22
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3If you understand French, there is a long discussion (partly numerical, partly theoretical) here – Giovanni Resta Feb 25 '16 at 8:28
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2Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions. – uvdose Feb 25 '16 at 8:37
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3@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$. – uvdose Feb 29 '16 at 14:25
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5If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps.... – 201044 Mar 15 '16 at 23:39
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.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
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.
