I tried to deny the thesis and show that $p$ is not prime. Like this, if $24\nmid p^{2}-1$ then $p^{2}-1=24q+r$, $0<r<24$, but it did not look so cool.
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What are the possible remainders $p$ might have $\pmod {24}$? – lulu Feb 28 '18 at 12:58
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First, it's well known that $p^2 \equiv 1 \pmod 8$. (Try letting $p = 8k + m$ for odd $m$ and compute the binomial expansion of $p^2$.)
Then, since $3 \nmid p$, $p \equiv \pm1 \pmod 3$, which implies $p^2 \equiv 1 \pmod 3$.
Finally, finish this question with Chinese Remainder Theorem: as $\gcd(3,8) = 1$, $p^2 = 1 \pmod{24}$ is equivalent to $p^2 \equiv 1 \pmod 8$ and $p^2 \equiv 1 \pmod 3$.
GNUSupporter 8964民主女神 地下教會
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No need to compute the binomial expansion : either $p \equiv 1, 3, 5 $ or $7 \pmod 8$, and the square of each of those residus is $\equiv 1 \pmod 8$. – krirkrirk Feb 28 '18 at 13:07
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@krirkrirk Yeah you're right, but for some secondary school students who haven't seen modular arithmetic, "binomial expansion" can convince them without introducing them new things. – GNUSupporter 8964民主女神 地下教會 Feb 28 '18 at 13:10
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This isn't introducing new things, since you're using in your answer $p \equiv \pm 1 \pmod 3 \Rightarrow p^2\equiv 1 \pmod 3$ – krirkrirk Feb 28 '18 at 13:12
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@krirkrirk I typed it like this since OP knows the "mod" notation. In my previous comment, I mean for some students who haven't learnt this notation, we can use $(8k+m)^2 = \cdots = 8(\star) + 1$ and $(3\ell \pm 1)^2 = \cdots 3(\star) + 1$ to show them these facts without the need to introduce new vocabulary. – GNUSupporter 8964民主女神 地下教會 Feb 28 '18 at 13:18
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If $p$ is prime and $p>3$ then $p \equiv1\pmod{4}$ or $p \equiv3\pmod{4}$ (otherwise $p$ is an even number, not true because $2$ is the only even prime number).
Because of that, either $n-1$ or $n+1$ is divisible by $4$ (but not both), the other number is also even (divisible by $2$), which implies $(n-1)(n+1)$ or $n^2-1$ is divisible by $8$.
Also if $p$ is prime and $p>3$ then $p$ is not divisible by $3$, so either either $n-1$ or $n+1$ is divisible by $3$ (but not both)$\Rightarrow n^2-1$ is divisible by 3.
Because $GCD(3;8)=1$, $n^2-1$ is divisible by $3 \times 8 =24$.
user061703
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If $p$ is prime we know that it must have odd residue.
But it can't be $\{3, 9, 15, 21\}$, as $24k + r$ with one of those is divisible by $3$.
This leaves $\{1, 5, 7, 11, 13, 17, 19, 23\}$, all of which when squared have remainder $1 \bmod 24$.
orlp
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The squares modulo $24$ are $0, 1, 4, 9, 12, 16$. There are no other possible values (see Sloane's OEIS).
If $n^2 \equiv 9 \pmod{24}$, that suggests $n$ is a multiple of $3$, and since you stipulated $p > 3$, this rules out $n = \pm 3$, the only primes divisible by $3$. And $0, 4, 12, 16$ are all even, but since you stipulated $p > 3$, $n = \pm 2$ is also ruled out.
That leaves us with $n^2 \equiv 1 \pmod{24}$. So if $p^2 \not \equiv 0, 4, 9, 12$ or $16$, what can it possibly be?
Look at the squares of a few small primes: $5^2 = 24 + 1$, $7^2 = 2 \times 24 + 1$, $11^2 = 5 \times 24 + 1$, etc.
David R.
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