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Prove that If gcd$(a,2)=1,$ and $p$ is an odd prime, then $(a^{p-1})-1$ is divisible by $8$.

amWhy
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Devansh
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    Welcome to Mathematics Stack Exchange. $a^n-1$ is divisible by $8$ for any even $n$ and odd $a$ – J. W. Tanner Mar 11 '20 at 20:36
  • Special case of a FAQ, for which we have at least $\color{#c00}{8\pm 1}$ proofs, and likely many more. – Bill Dubuque Mar 11 '20 at 20:45
  • Well. $p-1$ is even and $a$ is odd and $a^{p-1}-1=(a^{\frac {p-1}2}+1)(a^{\frac {p-1}2} -1)$ so...... – fleablood Mar 11 '20 at 20:46
  • ... and if you have to consecutive even numbers, (Such as $a^{k} +1$ and $a^k-1$ are if $a$ is odd), then one is divisible by $2$ but not $4$ and the other is divisible by $4$. (Because one of $m$ and $m+1$ is odd and the other is even so one of $2m$ and $2m+2$ is divisible by $2$ and one is divisble by $4$). – fleablood Mar 11 '20 at 21:11

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This is true if $p$ is ANY odd integer infact, $p$ does not have to be prime.

Note that $c^2 \equiv_8 1$ if $c$ is odd; it is easy to check, check $c \equiv_8 1,3,5,7$. Note that $a^{p-1}$ is an odd square for any odd $p$ and thus even $p-1$ i.e., $a^{p-1} = c^2$ for some odd $c$. So what must $a^{p-1}$ be mod 8. Can you finish from here.

Mike
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