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I suppose that $p-1$ is even number and that $p+1$ is divisible by 3 or vica versa. My first problem is that I needed to prove that $p^2 -q^2$ is always divisible by 24 for $p,q$ being prime numbers bigger than 3.

What I did was wrote $p^2 -q^2=(p-q)(p+q)$ ,given that $p$ and $q$ are prime numbers,they are not divisible by 2 .So I can write them as $2k +1$ ,$2m+1$).When I replace $p$ and $q$ into equation above I get $4(k-m)(k+m+1)$,which means that this polynomial is divisible by $4$. I just need to make sure that it is divisible by $6$ as well.Doing all $5k+1,2,3,4,5$ would be too long and I would need to check 25 cases.

So I wrote $p^2 - q^2=p^2 -1 - (q^2+1)=(p-1)(p+1) - (q-1)(q+1)$ If $p=2k+1 ,q=2m+1$ then polynomial is divisible by $4$.For testing with 3 I get that it is also divisible by 3.The most I can get is that it is divisible by 12,but how do I prove that it is divisible by 24,what am I missing?

Bill Dubuque
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Stephanie V
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    Hint: if $\gcd(a,3)=1$, then $a^2\equiv 1\pmod{3}$. If $a$ is odd, then $a^2\equiv 1\pmod{8}$. – Arturo Magidin Apr 13 '24 at 01:52
  • Does this answer your question? Suppose that $5\leq q\leq p$ are both prime. Prove that $24|(p^2-q^2)$. - found using an Approach0 search. – John Omielan Apr 13 '24 at 01:57
  • I see that the same question is asked before but I am aiming at better explanation – Stephanie V Apr 13 '24 at 02:04
  • It is obivious that if $p$ and $q$ are both odd then $p-q$ and $p+q$ are both even. But what is not obvious (but easy to show) is that if $p\pm q$ is divisible by $2$ but not divisible by $4$ then $p\mp q$ is divisible by $4$. (And conversily, if one is divisible by $4$ than the other is divisible by $2$ but not by $4$). And that's where thate mysterious extra $2$ factor comes from. – fleablood Apr 13 '24 at 03:02
  • "So I wrote p2−q2=p2−1−(q2+1)=(p−1)(p+1)−(q−1)(q+1)" That's not correct. $p^2-1= (p+1)(p-1)$ but $q^2 + 1\ne (q-1)(q+1)=q^2-1$. Why did you try to get these $1$s in? Just write it as $p^2 -q^2 = (p+q)(p-q)$. That's it. Then if $p =2k +1$ and $q=2m+1$ we get... $(p+q) = 2k+1 +2m + 1= 2k + 2m + 2$ and $p-q = 2k+1 -(2m +1) = 2k - 2m$ $(p+q)(p-q) =(2k + 2m + 1)(2k -2m) =4(k+m+1)(k-m)$. .... and do you see what is happening there? – fleablood Apr 13 '24 at 03:11
  • "When I replace p and q into equation above I get 4(k−m)(k+m+1),which means that this polynomial is divisible by 4" Exactly. And can you prove that $(k-m)(k+m+1)$ is divisible further by $2$? – fleablood Apr 13 '24 at 03:14

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I think that I partialy get it for $p=2k+1$,$(p-1)(p+1)=2k*2*(k+1)=4*k(k+1)$ which basically means that polynomial is divisible by 8 and we know that it is divisible by 3 when we test it,so it is divisible with 24.

Stephanie V
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  • These are not "polynomials", they are numbers. As a polynomial with integer coefficients, it is not divisible by $8$. – Arturo Magidin Apr 13 '24 at 01:57
  • Ok my mistake in semantics,but I think that point holds we have 4 as factors and two consecutive numbers,which means that one of them will be divisible by 2 – Stephanie V Apr 13 '24 at 02:06