Is $p^{2} \overset{8}{\equiv} 1$

Question

I was working on an assignment on finite fields, and I think it boiled down to prove the following, for every prime $p>2$:

$p^{2} \overset{8}{\equiv} 1$

I checked it directly for small primes, and it seems to hold. Is this statement true? If it is, I'd appreciate help in reaching a general proof.
Many Thanks!

In fact, for every prime $p>3$, $p^2=24k+1$, where $k$ is a generalized pentagonal number — Juan Moreno, Feb 10 '21 at 12:11

score 2 · Answer 1 · edited Feb 10 '21 at 12:19

Not only for prime numbers greater that 2, but also it's true for all odd numbers. For any odd integer $k$, the following is true: $k^2\overset{8}{\equiv}1$
The reason is that $k^2\overset{8}{\equiv}1 \iff 8\mid k^2-1=(k-1)(k+1)$
And because $k$ is odd, $(k-1),(k+1)$ are both even. Furthermore, one of two consecutive even numbers is also divisible by $4$ and since $(k-1),(k+1)$ are two consecutive even numbers, one of them is divisible by $2$ and the other is divisible by $4$ so their product is divisible by $8$.

TheSilverDoe · Accepted Answer · 2021-02-10T12:33:22.630

1

Actually, every odd number satisfies this property, since $$1^2=1 \equiv 1 \,[8]$$ $$3^2=9 \equiv 1 \,[8]$$ $$5^2=25 \equiv 1 \,[8]$$ $$7^2=49 \equiv 1 \,[8]$$

edited Feb 10 '21 at 12:33

answered Feb 10 '21 at 12:10

TheSilverDoe

29,720

1

OP asked for a proof. – Oussema Feb 10 '21 at 12:11
3

@Oussema This is a proof. – TheSilverDoe Feb 10 '21 at 12:12
You should clarify it. – Oussema Feb 10 '21 at 12:13
3

@Oussema I let the OP understand by himself/herself why checking only the odd numbers before $8$ is sufficient to deduce the property for all odd numbers. – TheSilverDoe Feb 10 '21 at 12:14
Fair enough, I guess. – Oussema Feb 10 '21 at 12:17
Please strive not to post more dupe answers to dupes of FAQs – Bill Dubuque Feb 10 '21 at 18:17

Is $p^{2} \overset{8}{\equiv} 1$

2 Answers2

Linked