If $p$ and $q$ are odd primes show that $$p|(9^q -1) \implies q|(p-1).$$ ....
I am having difficulty to prove this.
Yes that was the question I have misunderstood between 9 and q!!!
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why?? i cannot understand – User8976 Feb 22 '15 at 05:58
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1What you have is a factorization of $q^6-1$... Also, see my solution. – Calvin Lin Feb 22 '15 at 05:58
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sorry for that!!! i have gone mad!!! – User8976 Feb 22 '15 at 05:59
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1$19\mid 5^9-1$, but $5\nmid 19-1$. Also $19\nmid 5-1$. – Jyrki Lahtonen Feb 22 '15 at 06:01
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@tone note that Jykri's example shows you that $ p \not \mid q-1$ too. – Calvin Lin Feb 22 '15 at 06:04
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thanks for the counter examples – User8976 Feb 22 '15 at 06:05
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Probably it should be $\ p\mid 9^q-1,\Rightarrow, q\mid p-1\ $ – Bill Dubuque Feb 22 '15 at 06:16
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@tone what exactly is the correct question? Thanks – Jr Antalan Feb 22 '15 at 06:45
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Hint $ $ Apply twice: $\ 9^{\:\!\large n}\!\equiv 1\!\iff\! \color{#0a0}{{\rm ord}\,9}\mid n,\,$ by the Order Theorem, i.e.
$\!\!\bmod p\!:\ 9^{\:\!\large q}\equiv 1,\,$ so $\,9\,$ has $\rm\color{#0a0}{order}$ dividing $ $ prime $\,q,\,$ so it must be $\,\color{#0a0}q\,$ (not $\color{#c00}{\bf 1}$ else odd $\,p\mid 9^{\large \color{#c00}{\bf 1}}\!-1),\,$ thus by little Fermat $\,9^{\:\!\large p-1}\!\equiv 1,\,$ so $\,\color{#0a0}{q}\mid p\!-\!1.\ $ [note $\,p\mid 9^{\:\!\large q}\!-1\,\Rightarrow\,p\nmid 9,\,$ so Fermat applies].
Remark $ $ The same proof works if we replace $\,9\,$ by any integer $\,a\,$ and we add the hypothesis $\,p\nmid a^{\color{#c00}{\bf 1}}\!-1\,$ (to ensure $a$ has order $\,\rm\color{#0a0}{q}\,$ (vs. $\color{#c00}{\bf 1})$ modulo $\,p)$.
Bill Dubuque
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