Are $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime to each other, $n \in \mathbb{N}$?

Question

Let $p$ be a prime integer greater than $2$. Then I want to prove the followings:

$(1)$ $\frac{p^2+1}{2}$ and $\frac{p^5-1}{2}$ are coprime to each other.

$(2)$ $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime to each other, $n \in \mathbb{N}$

$(3)$ Are $\frac{p^2+1}{2}$ and $\frac{p^{5n-3}(p^5-1)}{2}$ coprime to each other, $n \in \mathbb{N}$ ?

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$(1)$

For $p=3$, $(p^2+1)/2=5$ and $(p^5-1)/2=141$ while $\gcd(5,141)=1$.

For $p=5$, $(p^2+1)/2=13$ and $(p^5-1)/2=1562$ while $\gcd(13,1562)=1$.

For $p=7$, $(p^2+1)/2=25$ and $(p^5-1)/2=8403$ while $\gcd(25,8403)=1$.

But how to prove the general case ?

From here, it is known that $a$ and $b$ are coprime if and only if $a^n$ and $b^n$ are coprime.

Thanks

Answer 1

This is for the question in the title. Let $q$ be a prime dividing both numbers. Then $q\mid p^2+1$, say $p^2+1 = Kq.$ Also note that $q\neq p$, so we must have $$q\mid p^5 -1 = p^5 + p^3 - p^3 - p +p -1 $$ $$= p^3(p^2+1) - p(p^2+1) + p-1$$

$$=p^3Kq -pKq + p-1.$$

Therefore $q\mid p-1$. And so $q\mid p^2-1$. Since it also divides $p^2+1$ it must divide $2$. So $q=2$.

But if $p$ is an odd prime then $p^2 = 4m+1$ for some $m$. So $(p^2+1)/2$ is odd, and so $2$ can't divide it.

Since the two numbers have no common factor, they are coprime.

Answer 2

Assume $\frac{p^2 + 1}{2}$ is not coprime with $\frac{p^{5n}(p^5 - 1)}{2}$, and that $\frac{p^2 + 1}{2} < \frac{p^{5n}(p^5 - 1)}{2}$, so then $\frac{p^2 + 1}{2} \mid \frac{p^{5n}(p^5 - 1)}{2}$.

which means that $\exists q \in \mathbb{N}$ such that:

$$\frac{p^{5n}(p^5 - 1)}{2} = \frac{p^2 + 1}{2} q$$

Then,

$$q = \frac{p^{5n}(p^5 - 1)}{p^2 + 1}$$

This means, $(p^2 + 1) \mid p^{5n}$ and/or $(p^2 + 1) \mid (p^5 - 1)$

So, $\frac{p^{5n}}{(p^2 + 1)} = p^{5n - 2} - p^{5n - 4} - p^{5n - 6}...$

If n is even, there will be a remainder value of $-\frac{1}{p^2 + 1}$, so $(p^2 + 1) \nmid p^{5n}$. If n is odd, $(p^2 + 1) \nmid p^{5n}$.

So therefore, $(p^2 + 1) \mid (p^5 - 1)$.

However, $(p^5 - 1) = (p^2 +1)(p^3 - p) + (some\ remainder)$.

This means, $(p^2 +1) \nmid p^{5n}$ and $(p^2 + 1) \nmid (p^5 -1)$.

So, $q = \frac{p^{5n}(p^5 - 1)}{(p^2 + 1)}$ is irreducible, and so $q \notin \mathbb{N}$.

We've arrived at a contradiction, and so therefore our original assumption is false, and therefore, both $\frac{p^2 + 1}{2}$ and $\frac{p^5 -1}{2}$ are coprime.

[Addendum] However, as pointed out by @MAS, what if they have a common factor > 1. So, again, assume both $\frac{p^2 + 1}{2}$ and $\frac{p^{5n}(p^5 - 1)}{2}$ are not coprime and suppose $\exists q,r,s \in \mathbb{N}$ such that $s$ is this common factor and $gcd(q, r) = 1$:

$$\frac{p^2 + 1}{2} = qs$$ $$\frac{p^{5n}(p^5 - 1)}{2} = rs$$

Then,

$$p = \sqrt{2qs - 1}$$

Plugging this into $\frac{p^{5n}(p^5 - 1)}{2}$, we get:

$$ \frac{\sqrt{2qs - 1}^{5n}(\sqrt{(2qs - 1)^5} - 1)}{2}$$

Then as stated, $$\frac{\sqrt{(2qs - 1)^{5n}}(\sqrt{(2qs - 1)^5} - 1)}{2} = rs$$

So,

$$\sqrt{(2qs - 1)^{5n}}(\sqrt{(2qs - 1)^5 - 1)} = 2rs$$

$$(2qs - 1)^{\frac{5n + 5}{2}} - (2qs - 1)^{\frac{5n}{2}} = 2rs$$

Since $s \mid 2rs$, then $s \mid (2qs - 1)^{\frac{5n + 5}{2}} - (2qs - 1)^{\frac{5n}{2}}$

So $s \mid (2qs - 1)^{\frac{5n}{2}}$. This means $\exists v \in \mathbb{N}$ such that

$$vs = (2qs - 1)^{\frac{5n}{2}}$$

Squaring both sides:

$$ (vs)^2 = (2qs - 1)^{5n}$$

$$ (vs)^2 = \sum_{i=0}^{5n}{5n \choose i}(2qs)^{5n-i}(-1)^i$$

$$ (vs)^2 = \sum_{i=0}^{5n - 1}{5n \choose i}(2qs)^{5n-i}(-1)^i + 1$$

While $s \mid (vs)^2$, $s \nmid \sum_{i=0}^{5n - 1}{5n \choose i}(2qs)^{5n-i}(-1)^i + (-1)^{5n}$ because while $s \mid \sum_{i=0}^{5n - 1}{5n \choose i}(2qs)^{5n-i}(-1)^i$, $s \nmid (-1)^{5n}$.

Therefore since $$s \nmid (2qs - 1)^{\frac{5n}{2}}$$

$$s \nmid (2qs - 1)^{\frac{5n + 5}{2}} - (2qs - 1)^{\frac{5n}{2}}$$

And so, since $$s \mid \frac{p^2 + 1}{2}$$, but $$s \nmid \frac{p^{5n}(p^5 - 1)}{2}$$

Therefore, $\nexists s \in \mathbb{N}$ such that $s > 1$ and

$$ s \mid \frac{p^2 + 1}{2}$$ and $$ s \mid \frac{p^{5n}(p^5 - 1)}{2}$$

So,

$$gcd(\frac{p^2 + 1}{2}, \frac{p^{5n}(p^5 - 1)}{2}) = 1$$

Therefore, $\frac{p^2 + 1}{2}$ and $\frac{p^{5n}(p^5 -1)}{2}$ are coprime.

NB: Corrections are appreciated.