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If $p$ and $q$ are two primes greater than $3$ then prove that $24\mid p^2-q^2$ . My solution goes like this:

If $p,q$ are primes greater than $3$ then $8 \mid p^2-q^2$ as any odd perfect square is of the form $8k+1$ where $k\in\mathbb{Z}$ . Now, $p\equiv 1,2 \pmod{3}$ and $q\equiv 1,2\pmod{3}$ ($0 $ cant be a case as $p,q$ are primes). So, we see that all possible combinations in the remainders of $p,q$ will be equivalent to zero either by doing $p+q$ or by $p-q$ such that $3 \mid p^2-q^2$ . So, $24 \mid p^2-q^2$.

I just want to verify whether my solution is okay or not.

Andreas Caranti
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    The first part, about the divisibility by $8$, is ok. As to the second part, just note that $p^{2} \equiv 1 \equiv q^{2} \pmod{3}$. – Andreas Caranti Apr 12 '22 at 10:36
  • @AndreasCaranti Yeah but then again $ p^2-q^2\equiv 0(mod 3) $ holds true....so that shouldn't be a problem .... –  Apr 12 '22 at 11:15
  • This question has appeared many times on this site...the duplicate I linked to is itself a duplicate. – lulu Apr 12 '22 at 11:52
  • @lulu No..actually i wanted to verify my solution ... –  Apr 12 '22 at 11:53
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    As I think you'll see, your argument matches the arguments that appear in the duplicates. – lulu Apr 12 '22 at 11:57

2 Answers2

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$p^2,q^2 \mod 24 \equiv 1$ for $p,q>3$

since $p^2 =(p-1)(p+1)+1$ where both of $(p-1)$ and $(p+1)$ are divisible by $2$, either one is divisible by $3$ and one by $4$ respectively.

Thus, their difference $p^2-q^2 \mod 24 \equiv 0$.

starrin
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Your solution is right, but by dividing your solution with parts, you can arrange it so you can read it easily.

\begin{align} & \text{STEP 1. } 3 | p^2-q^2. \\ & p, q > 3 \Rightarrow p^2 \equiv q^2 \equiv 1(\text{mod } 3). \\ \therefore \; & p^2-q^2 \equiv 0(\text{mod } 3). \\ \ \\ & \text{STEP 2. } 8|p^2-q^2. \\ & p, q > 3 \Rightarrow p, q: 8k\pm1 \text{ or } 8k\pm 3. \\ \therefore \; & p^2 \equiv q^2 \equiv 1 (\text{mod } 8). \\ \therefore \; & p^2-q^2 \equiv 0(\text{mod } 8). \\ \ \\ \therefore \; & 24 | p^2-q^2. \end{align}

RDK
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