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I asked this question earlier and asked people to not tell me the answer. After a few more hours of attempts, I cannot figure this out. Can someone please tell me how to solve this proof?

Suppose $(6,a) = (6,b) = 1$. Prove $24 \mid (a^2-b^2)$.

John Mancini
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  • In fairness to the new contributor, he asks about a class of numbers more expansive than the primes $\ge 5$, even though the arguments that apply to those primes apply to his class of numbers (because those primes are contained in his class of numbers). So the 'duplicate' designation is being applied perhaps overly strictly here. – Keith Backman Sep 07 '19 at 15:36

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$a$ and $b$ are both odd. That tells you something about how even the difference of their squares must be.

Each of $a$ and $b$ is congruent to $1$ or $2$ module $3$. That tells you something about the difference of their squares modulo $3$.

Ethan Bolker
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  • Very simple, very nice! ----- (It's possible to help a bit more in the second part by saying: [...] about their squares and about the difference of their squares mod 3). – Wlod AA Sep 07 '19 at 04:11
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Since we know that $a$ and $b$ are coprime to 6, let's reframe the problem in mod 6. Since $(6, a) = (6, b) = 1$, neither 2 nor 3 can divide $a$ or $b$. Thus $a$ and $b$ must be $1$ or $5$ mod 6, or equivalently $1$ or $-1$ mod 6. This means that we can write $a = 6m\pm 1$, $b = 6n\pm 1$ for some integers $m$, $n$. Noting that $$a^2 - b^2 = (a+b)(a-b)$$ and substituting the expressions for $a, b$ in terms of $m, n$ should give you what you need.

Andrew
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