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Let $\gcd(b,6) = 1$. Prove that $b^2 \equiv 1 $ (mod 24).

Now I have that as $\gcd(b,6) = 1$, we know that $3\nmid b $ and $2\nmid b$ (else the GCD would be 3 or 2 resp.)

So as $2\nmid b$, $b$ must be odd. Hence $b^2$ is also odd.

Then I'm not sure where to take it from here? Maybe I've gone down the wrong path to start with, and should be looking more along the lines of the Chinese Remainder Theorem? I'm not sure.

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  • $\begingroup$ Since $b^2$ is a perfect square, we have that $b^2 = 1,4,9,16$ modulo $24$. All of these, besides $1$, require that $2 | b$ or $3 | b$. $\endgroup$ – Eric Thoma Apr 9 '14 at 19:13
  • $\begingroup$ Thank you! this is a very nice simple trick. $\endgroup$ – jakey Apr 9 '14 at 19:28
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Starting with $b$ is odd, so $b = 2k + 1$, and $b^2 - 1 = 4k^2 + 4k = 4k(k+1)$ clearly divisible by $8$. So we need to show that $b^2 - 1$ is divisible by$3$. Since $b$ is not divisible by $3$, $b = 3k + 1$ or $b = 3k +2$. If $b = 3k + 1$, then $b - 1 = 3k$ and $3 | b - 1$, so $3 | b^2 - 1$. And if $b = 3k + 2$, then $b + 1 = 3k + 3$ which is divisible by $3$. So $3 | (b +1 )(b - 1) = b^2 - 1$. So $3 | b^2 - 1$ and since $gcd(3,8) = 1$, $24 | b^2 - 1$.

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  • $\begingroup$ Thank you! I had previously shown if b is odd then $b^2 \equiv 1$ (mod 8) so I feel as this is incorporated in to your anwser it bodes well! $\endgroup$ – jakey Apr 9 '14 at 19:27
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Hint $\rm\ 3\nmid b,\,$ so $\rm\, mod\ 3\!:\ b\,\equiv\, \pm1\ \ \Rightarrow\ \ b^2\equiv 1$

$\rm\ \ and \ \ 2\nmid b,\,$ so $\ \rm mod\ 8\!:\, b\equiv \pm1,\pm3\,\Rightarrow b^2\equiv1,\ $ so $\rm\,\ 3,8\mid b^2\!-\!1 \Rightarrow\, 3\cdot 8\mid b^2\!-\!1$

Remark $\ $ More generally see the Carmichael function.

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