<s>[INST] Determine all positive integers $n$ such that $n^2+1$ divides $2^{n!}-1$ For this question, I have tried to start from reading similar question (How many rationals of the form $\frac{2^n+1}{n^2}$ are integers? from this post How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers?)
I notice that $2^{2!}-1$ is always odd so I tried to rewrite it as $2k+1$ but I am stuck afterward. I am thinking of how do I relate this to modular arithmetic or gcd. I would appreciate if anyone could help.
 [/INST] In the link in comment it is proved that $m^2-1\big|2^{m!}-1$ . Let $m=n^2$, then $m^2-1=n^4-1\big|2^{m!=n^2!}-1$ and $n^2+1\big|(n^4-1)$. For example :
$n=4 \rightarrow 4^2+1=17\big|4^4-1$ and we have:
$2^{4!}-1=2^{24}-1=[(2^4)^3-1][(2^4)^3+1]$
$(2^4)^3+1=(2^4+1)[(2^4)^2-(2^4)+1]=17k$
therefore:
$4^2+1\big|2^{4!}-1$
In fact $n=4$ is the smallest number and general form of n is $2t$ such that $(2t)^2+1$ is a prime. Another example:
$10^2+1=101\rightarrow 2^{101}\equiv 1 \bmod 101$, $100!=100(s)$ and $2^{100!}=(2^{100})^s\equiv 1 \bmod 101\Rightarrow 10^2+1\big|2^{100!}-1$.
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