If $(6,n)=1$, prove that $n^2-1$ is divisible by $24$.
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Okay, what are your thoughts on this so far? What exactly is your question? – Théophile Nov 19 '18 at 19:16
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I assume this means $\gcd(6,n)=1$? – Kevin Long Nov 19 '18 at 19:31
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Most (if not all) of the proofs in the dupe apply here (even though the title restricts to primes $> 3)\ $ – Bill Dubuque Nov 19 '18 at 20:21
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Hint
$$(6,n)=1\implies$$ $ n=6k+1 $ or $n=6k-1$.
in the first case
$$n^2-1=36k^2+12k=12k(3k+1)$$
in the second
$$n^2-1=12k(3k-1).$$
If $k$ is even ....
hamam_Abdallah
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https://math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511 – lab bhattacharjee Nov 19 '18 at 20:10
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If $(6,n) = 1,$ then $n \equiv 1 \pmod 6$ or $n \equiv 5 \pmod 6,$ since $\varphi(6) = 2.$ So, we have two cases:
► $n \equiv 1 \pmod 6:$ $$n \equiv 1 \pmod 6 \Rightarrow n^2 \equiv 1^2 \pmod 6 \Rightarrow n^2 - 1 \equiv 0 \pmod 6 \Rightarrow \boxed{n^2 - 1 \equiv 0 \pmod{24}}$$
► $n \equiv 5 \pmod 6 \Rightarrow n \equiv -1 \pmod 6:$ $$n \equiv -1 \pmod 6 \Rightarrow n^2 \equiv (-1)^2 \pmod 6 \Rightarrow n^2 - 1 \equiv 0 \pmod 6 \Rightarrow \boxed{n^2 - 1 \equiv 0 \pmod{24}}$$
In both, we conclude that $n^2 - 1 \equiv 0 \pmod{24}.$ Therefore, $n^2 - 1$ is divisible by $24$ when $(6,n) = 1.$
674123173797 - 4
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$(6,n)=1$ $\implies$ $n$ is odd. Therefore , $$n^{2}\equiv 1 \pmod8$$ Also $(3,n)=1$ , since $(2\times3,n)=1$ Therefore by Fermat's theorem ,$$n^2\equiv 1\pmod3$$. Now you can directly combine these two congruences, since $(3,8)=1$
Therefore, $$n^2\equiv 1\pmod{24}$$