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Let p be an odd prime and suppose b is an integer whose order mod p = 7. Show that -b has order 14.

Here is where I am at:

I can rewrite then b^7 congruent to 1 mod p

Now, collary 7.2 in my book states order of a mod p divides p-1 .

So then 7|p-1. I am kind of stuck here...and I am looking for feedback on how to approach this using a valid theorem or proposition, because the one I am using is not going too well.

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The order of $b$ module $p$ being equal to $7$ implies, by Lil' Fermat, that $7$ is a divisor of $p-1$, i.e. $p\equiv 1\mod 7$ (and indeed $p\equiv 1\mod 14$ since $p$ is odd).

Now $-1$ has order $2$ modulo any odd prime, hence $-b=(-1)b$ has order $\operatorname{lcm}(2,7)$.

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Clearly, $(-b)^{14} \equiv b^{14} \equiv 1$ and so the order of $-b$ is a divisor of $14$.

$(-b)^{1} \equiv -b \not\equiv 1$ and so its order is not $1$.

$(-b)^{2} \equiv b^2 \not\equiv 1$ and so its order is not $2$.

$(-b)^{7} \equiv -1 \not\equiv 1$ and so its order is not $7$.

So the order must be $14$.

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  • $\begingroup$ I don't understand why you stated Clearly, (-b)^14 is congruent to one..I thought we don't know this. This is what we are trying to prove. $\endgroup$
    – JD112
    Nov 15 '16 at 0:09
  • $\begingroup$ @JD112, we are trying to prove that $14$ is the smallest exponent that works. $\endgroup$
    – lhf
    Nov 15 '16 at 0:10
  • $\begingroup$ Ah okay I think I see. Lastly, can we say (-b)^7 is congruent to -1 because b^7 is congruent to 1? $\endgroup$
    – JD112
    Nov 15 '16 at 0:11
  • $\begingroup$ @JD112, yes.... $\endgroup$
    – lhf
    Nov 15 '16 at 0:12

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