Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime 
Let $p>3$ be a prime number.  How to prove that $\left(p^2+1\right)^2$ has no divisors congruent to $1 \bmod p$, except the trivial ones $1$, $p^2+1$, and $\left(p^2+1\right)^2$? 

When $p=3$, you also have $p+1$ as a divisor of $\left(p^2+1\right)^2$.


Proposition.  For a positive integer $p$, $\left(p^2+1\right)^2$ is divisible by some positive integer $d\equiv 1\pmod{p}$ such that $$d\notin\left\{1,p^2+1,\left(p^2+1\right)^2\right\}$$ if and only if
  $$p=t_s(b)\text{ for some integers }b\geq 1\text{ and }s\geq 2\,,$$
  where
  $$t_j(b):=\frac{\left(\frac{\left(b^2+2\right)+b\,\sqrt{b^2+4}}{2}\right)^j-\left(\frac{\left(b^2+2\right)-b\sqrt{b^2+4}}{2}\right)^j}{\sqrt{b^2+4}}$$
  for all positive integers $b$ and $j$.    Additionally, if $p$ is in this form $t_s(b)$, then there are at least two positive integers not of the form $1$, $p^2+1$, and $\left(p^2+1\right)^2$ that divide $\left(p^2+1\right)^2$, which are
  $$t_s(b)\,t_{s-1}(b)+1\text{ and }\left(b^2+2\right)\,\left(t_s(b)\right)^2-t_s(b)\,t_{s-1}(b)+1\,.$$
  Finally, the only prime natural number $p$ with this property is $p=3$.

Suppose that $p$ and $d$ are positive integers such that $d\equiv 1\pmod{p}$ and $d$ divides $\left(p^2+1\right)^2$.  Without loss of generality, we may assume that $d\leq p^2+1$; otherwise, replace $d$ by $\dfrac{\left(p^2+1\right)^2}{d}$.  Because $d\equiv 1\pmod{p}$, we have
$$d=pk+1\text{ for some integer }k\text{ such that }0\leq k\leq p\,.$$
As $d\mid \left(p^2+1\right)^2$, we get that $d=pk+1$ divides
$$\left(p^2+1\right)^2+(pk+1)(-2p^2+pk-1)=p^2\,\left(p-k\right)^2\,.$$
Since $\gcd\left(pk+1,p^2\right)=1$, it follows that 
$$\frac{(p-k)^2}{pk+1}\in\mathbb{Z}\,.$$
Of course, $k=0$ and $k=p$ are possible solutions, but they lead to trivial divisors $d=1$ and $d=p^2+1$.  
From Zvi's answer in this link, all integral solutions $(p,k)$ with $p>k>0$ take the form
$$(p,k)=\big(t_{j+1}(b),t_j(b)\big)\text{ for some positive integers }b\text{ and }j\,,$$
where $\left\{t_j(b)\right\}_{j\in\mathbb{Z}_{\geq 0}}$ is a sequence defined by the initial conditions $t_0(b):=0$ and $t_1(b):=b$, along with the recurrence 
$$t_{j}(b):=\left(b^2+2\right)\,t_{j-1}(b)-t_{j-2}(b)\text{ for all integers }j\geq 2\,.$$
For a given positive intger $b$, note that
$$t_j(b)=\frac{\left(\frac{\left(b^2+2\right)+b\,\sqrt{b^2+4}}{2}\right)^j-\left(\frac{\left(b^2+2\right)-b\sqrt{b^2+4}}{2}\right)^j}{\sqrt{b^2+4}}\text{ for all }j=0,1,2,\ldots\,.$$
From here, we conclude that all positive integers $p$ for which $\left(p^2+1\right)^2$ is divisible by some positive integer $d\equiv 1\pmod{p}$ not of the form $1$, $p^2+1$, or $\left(p^2+1\right)^2$ must satisfy
$$p=t_{s}(b)\text{ for some integers }b>0\text{ and }s>1\,.$$
Amongst $1$ to $100$, positive integers $p$ in the said form are
$$t_2(1)=3\,,\,\,t_3(1)=8\,,\,\,t_2(2)=12\,,\,\,t_4(1)=21\,,\,\,t_2(3)=33\,,$$
$$t_5(1)=55\,,\,\,t_3(2)=70\,,\text{ and }t_4(1)=72\,.$$
For example, with $p:=t_2(2)=12$, we have $\left(p^2+1\right)^2=145^2$, which is divisible by $$d:=5^2=25\equiv 1\pmod{p}\,,$$ which is not of the form $1$, $p^2+1$, or $\left(p^2+1\right)^2$.  In particular, if $p$ is a prime natural number, then we must have
$$p=t_2(1)=3\,,$$
where $\left(p^2+1\right)^2=10^2=100$ has a divisor $d:=4\equiv 1\pmod{p}$ not of the form $1$, $p^2+1$, or $\left(p^2+1\right)^2$.
