Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime

Question

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 $\left(p^2+1\right)^2$.

Answer 1

I think this is a good attempt at an answer, but I'm not 100% confident. I'll put it out there anyway. I start by looking at the factors of $(p^2+1)$. For $a_i,b_j\in \mathbb{N}$, let:

$$2\le a_i<(p-1)<(p+1)<b_j\le \frac{p^2+1}{2}$$ For certain $a_k,b_k$, $a_kb_k=(p^2+1)$. $a_k\equiv a_k\not\equiv 1\mod{p}$. Hence, $b_k\not\equiv 1\mod{p}$. Rather, $b_k\equiv \bar a_k\mod{p}$ where $\bar a_k$ represents the multiplicative inverse of $a_k\mod{p}$.

Therefore, $(p^2+1)$ has no factors of the form $(np+1)$. This should mean that $(p^2+1)^2$ has also no such factors.

Assume there is a factor $q$ of $(p^2+1)^2$ which is larger than $(p^2+1)$. If $q\equiv 1\mod{p}$, then there should be another factor $r=\frac{(p^2+1)^2}{q}<(p^2+1)$ where $r\equiv 1\mod{p}$. But that is not possible.

In sum (unless I have made an error), there are no factors of $(p^2+1)^2$ that are congruent to $1\mod{p}$.

Answer 2

Assume there are factors of $\def\num{(p^2+1)^2}\num$ which are congruent to $1$ modulo $p$. Then both the factor and the quotient are $\equiv1\pmod p$ as $\num\equiv1\pmod p$, so we can write $(np+1)(mp+1)=\num$.

Then $$ (nmp+n+m)p+1=p^4+2p^2+1, $$ so $nmp+n+m=p^3+2p$. This shows that $p$ divides $n+m$.

Without loss of generality, write $n=a_np+k$ and $m=a_mp-k$ where $a_m\geq1$ and $0\leq k<p$.

Substituting in the equation $nmp+n+m=p^3+2p$, we obtain $$a_na_mp^2+(a_m-a_n)kp-k^2+a_n+a_m=p^2+2.$$

Now $(np+1)(mp+1)=(p^2+1)^2$ means one of $n$ and $m$ is $\leq p$, otherwise $(np+1)(mp+1)>\num$. This implies we only have to consider three cases: $a_n=1,\,k=0$, $a_n=0$ or $a_m=1$.

$a_n=1,\,k=0$:

The equation becomes $$a_mp^2+a_m=p^2+1.$$ This means $a_m=1$, and we obtain the solution $(p^2+1)(p^2+1)=(p^2+1)^2$.

$a_n=0$:

The equation becomes $$a_mkp-k^2+a_m=p^2+2.$$ If $a_m>p$, then the left-hand side is larger than the right-hand side of the equation, so $a_m\leq p$.

Since $a_m-k^2\leq p$, we must have $a_m=p$ and $k=1$. It follows that $a_m-k^2=2$, so $p=3$. And we obtain the case $(p+1)((p^2-1)p+1)=(p^2+1)^2$ for $p=3$.

$a_m=1$:

Write the equation as $$((a_np+k)p+1)((p-k)p+1)=p^4+2p^2+1.$$ Hence $$(a_np+k)(p-k)p^2+(a_np+p)p+1=p^4+2p^2+1$$ and $$(a_np+k)(p-k)+a_n=p^2+1.$$ Namely, $$a_n(p^2-kp+1)=p^2+1-kp+k^2.$$

This means that $p(p-k)+1$ divides $k^2$. Note that here $k\ne0$, otherwise it is just the case $(p^2+1)(p^2+1)=(p^2+1)^2$. Substituting $k$ by $p-k$, we obtain $pk+1\mid(p-k)^2$. By this question, this is impossible.

Therefore the factors of $\num$ which are $\equiv1\pmod p$ are as described.

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Hope this helps. Perhaps I shall post this as community wiki, as it involves an answer from other users.

Answer 3

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 $\left(p^2+1\right)^2$.