How to solve these two simultaneous “divisibilities” : n+1∣m2+1n+1\mid m^2+1 and m+1∣n2+1m+1\mid n^2+1

Is it possible to find all integers m>0 and n>0 such that n+1m2+1 and m+1|n2+1 ?

I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.

Thanks !

Answer

Some further results along the lines of thought of @individ:

Suppose p and s are solutions to the Pell’s equation:
dp2+s2=1
Then,
m=ap2+bpq+cq2n=ap2bpq+cq2
are solutions if (a,b,c,d) are: (these are the only sets that I found using the computer)
(10,4,2,15)(39,12,3,65)
Sadly, the solutions are negative.

Here are some examples:
(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)
P.S. I am also very curious how @individ thought of this parametrization.

Attribution
Source : Link , Question Author : uvdose , Answer Author : Yifan Zhu

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