8

Let $n$ be a positive integer. Show that if $p$ is a prime dividing $1-5^n+5^{2n}-5^{3n}+5^{4n}$, then $p\equiv 1 \pmod{4}$.

Here is what I have tried: let $r=-5^n$, then $1-5^n+5^{2n}-5^{3n}+5^{4n}=1+r+r^2+r^3+r^4=\frac{1-r^5}{1-r}$. Hence $ord_p(1-r^5)-ord_p(1-r)>0.$ Now clearly $p\not= 2,5$, so $5^{(p-1)/2}\equiv -1\pmod{p}$. So $ord_p(1-r)=ord_p(1+5^n) | (p-1)/2$.

Similarly, $1-r^5=1+5^{5n}$ and $(5^{5n})^{(p-1)/2}\equiv -1 \pmod{p}$ but then I don't know how to make of that.

I see that this problem is about $p\equiv 1 \pmod{4}$ so I thought it might have something to do with the Gaussian integers $\mathbb{Z}[i]$ but I am not sure how the theory of gaussian integers helps in this case. And the number $5$ and the exponent $4$ seem suspicious, can we generalize it and replace them with other prime $q$ and $q-1$?

Thanks in advance.

Ishigami
  • 2,202
  • 3
    Your calculations show that $ord_p(5^n)=10$. This implies $10\mid p-1$. Based on numerical testing (aided by Mathematica) it seems to me that we actually have $20\mid p-1$. If we could show that $5$ is a quadratic residue modulo $p$, say $5\equiv a^2$, it would follow that $a^n$ has order $20$ modulo $p$ and you would be done. We have looked at similar problems earlier on this site. But it's past midnight, and I'm too tired to search for the closest match right now. – Jyrki Lahtonen Jan 16 '25 at 22:24
  • 1
    (cont'd) Are you familiar with quadratic reciprocity? Can you complete the argument from my first comment? I actually suspect this is a duplicate. A better known case of a similar argument is here. – Jyrki Lahtonen Jan 16 '25 at 22:27
  • 2
    @JyrkiLahtonen From quadratic reciprocity, $5$ is a quadratic residue iff $p \equiv 1,4 \pmod{5}$. However, with $\operatorname{ord}_p(5^n)=10$, then $p\equiv 1\pmod{5}$, as you've indicated, so $5$ is a quadratic residue, say $5\equiv r^2\pmod{p}$. Then $5^{5n}\equiv -1\pmod{p};\to;(r^{5n})^2\equiv -1\pmod{p}$, so $-1$ is a quadratic residue modulo $p$, i.e., $p\equiv 1\pmod{4}$. I've written an answer using this, but was hesitant to post it, since you're right it's likely some people will consider it a duplicate. Anyway, it's a moot point since quasi has already posted a similar answer. – John Omielan Jan 17 '25 at 01:34
  • It can be shown that $(\frac {a^m-1}{a-1}, a-1)=(a-1, m)$. What you found means $a=r$ and $m=5$so we have a common divisor like $d=(r-1, 5)$. So the trivial common divisor can be $5$ which can be taken as the prime dividing the progression. and we have $p=5\equiv 1\bmod 4$. – sirous Jan 18 '25 at 11:37

1 Answers1

4

Let $n$ be a positive integer, and let $p$ be a prime divisor of $$ 1-5^n+5^{2n}-5^{3n}+5^{4n} $$ Then $p$ must be odd, and $p\ne 5$.

From the identity $$ 1-5^n+5^{2n}-5^{3n}+5^{4n} = \frac {5^{5n}+1} {5^n+1} $$ it follows that $5^{5n}\equiv -1\;(\text{mod}\;p)$.

Suppose $5{\,\not\mid\,}p-1$.

Let $e$ be the order of $5$, mod $p$.

Necessarily $e{\,\mid\,}p-1$. \begin{align*} \text{Then}\;\;& 5^{5n}\equiv -1\;(\text{mod}\;p) \\[4pt] \implies\;& 5^{10n}\equiv 1\;(\text{mod}\;p) \\[4pt] \implies\;& e{\,\mid\,}\gcd(p-1,10n) \\[4pt] \implies\;& e{\,\mid\,}\gcd(p-1,2n) \\[4pt] \implies\;& e{\,\mid\,}2n \\[4pt] \implies\;& 5^{2n}\equiv 1\;(\text{mod}\;p) \\[4pt] \implies\;& 5^n\equiv \pm 1\;(\text{mod}\;p) \\[4pt] \implies\;& 1-5^n+5^{2n}-5^{3n}+5^{4n}\equiv 1\;\text{or}\;5\;(\text{mod}\;p) \\[4pt] \implies\;& 0\equiv 1\;\text{or}\;5\;(\text{mod}\;p) \\[4pt] \end{align*} contradiction.

Hence $5{\,\mid\,}p-1$.

Then $p\equiv 1\;(\text{mod}\;5)$, so $p$ is a quadratic residue mod $5$.

By quadratic reciprocity, it follows that $5$ is a quadratic residue mod $p$.

Then we get \begin{align*} & 5^{\frac{p-1}{2}}\equiv 1\;(\text{mod}\;p) \\[4pt] \implies\;& \left(5^{\frac{p-1}{2}}\right)^{5n}\equiv 1\;(\text{mod}\;p) \\[4pt] \implies\;& \left(5^{5n}\right)^{\frac{p-1}{2}}\equiv 1\;(\text{mod}\;p) \\[4pt] \implies\;& (-1)^{\frac{p-1}{2}}\equiv 1\;(\text{mod}\;p) \\[4pt] \implies\;& \frac{p-1}{2}\;\text{is even} \\[4pt] \implies\;& p\equiv 1\;(\text{mod}\;4) \\[4pt] \end{align*} as was to be shown.

quasi
  • 61,115