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Given $p$ a prime number such that $p \geq 5$

Show that $p^2$ is congruate to $1$ modulo $24$

This is what I tried :

We have $p^2$ congruate to 1 modulo $3$ because if $p=3k+2$ so p is congruate to -1 modulo 3 so p^2 is congruate to 1 modulo 3

And $p^2$ is congruate to $1$ modulo $8$ because $p=2k+1$ ($p\geq5$) so $p^2-1=4q(q+1)=8q'$

We have $p^2=3k+1$ and $p^2=8k'+1$

So $p^4=24kk'+3k+8k'$ and $3k=p^2-1$ and $8k'=p^2-1$ So :

$p^4-2p^2+2=24kk'$

Im stuck here ! I don't want use to use a proprietie directly

DeepSea
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user233658
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    Your work is needlessly complicated. Note $p^2-1=(p+1)(p-1)$. But $p$ is prime, so $p-1$ and $p+1$ are even; in fact, one is a multiple of 4, because every other even number is divisible by 4. Hence $(p-1)(p+1)$ is divisible by 8. It must also be divisible by 3 because one of $p-1$, $p$, and $p+1$ is divisible by 3, but it's not $p$, because $p$ is a prime bigger than 3. – symplectomorphic Mar 23 '16 at 21:38

2 Answers2

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Prime numbers greater than 3 can always be expressed as $(6n+1)$ or $(6n-1)$ when we square a number in that form we get $(36n^2 \pm 12n + 1)$ and $(36n^2 \pm 12n + 1)\equiv 1(\mod 24)$

Doug M
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You're almost there.

You have proved that $p^2-1$ is a multiple of both $3$ and $8$ and so it must be multiple of their lcm, which is $24$.

lhf
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