<s>[INST] Generalisation of IMO 1990/P3:For which $b $ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$? 
For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?

It was from my LTE/ Zsigmondy handout.
By taking examples, it looks like for $b= 2^k-1 , 2$ it's not true .
Here's my progress:

*

*I got $b=4,5,6,8,9$ works ( $2,3,7$ doesn't )

*$n$ is odd : If not then $4\mid b^n-1$ , but $b^n \equiv 0,1 \mod 4$

*If $b+1$ is a power of $2$ and $n^2\mid b^n+1$, then $n=1$ : Let $p$ smallest prime dividing $n$ ( note that it can't be $2$) . Then $n^2\mid b^{2n}-1 \implies p\mid b^{2n}-1 \implies p\mid b^{\gcd(2n,p-1)}-1 \implies p\mid b^2-1\implies p|b-1 \text{(since $b+1$ is power of $2$)}\implies p\mid b^n+1 \implies p|2.$ Contradiction.

Any hints? Thanks in advance!
 [/INST] We can construct b, we want:
$b^n+1\equiv 0 \mod (n^2)$
If $b=t.n^2-1$ and $n =2k+1$ is odd, then we have:
$(t.n^2-1)^n+1=M(n^2)-1+1=M(n^2)\equiv 0 \ mod (n^2)$
Where $M(n)$ is a multiple of $n^2$.
The condition is that n must be odd.
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