I am interested in solutions to this congruence:
$$2^kx \equiv x \bmod m$$
Where $m$ and $x$ are known positive integers. They may not necessarily be prime or coprime. I am looking for solutions for $k$. How can I do this?
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1How about the cancellation law to cancel the $x$s first? – meiji163 Jun 20 '15 at 03:49
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1I don't know the cancellation law, or if it's even valid, because dividing both sides by $x$ would mean it would need to have a multiplicative inverse modulo $m$ which may not always be true – Remy Hartfield Jun 20 '15 at 03:53
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Here is a formal proof https://proofwiki.org/wiki/Common_Factor_Cancelling_in_Congruence – meiji163 Jun 20 '15 at 04:00
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If, in the terms the proof uses $x=a=b\equiv x (\text{mod} m)$ then $2^{k}\equiv1 (\text{mod} {m \over d})$ where $d = \gcd \left{{a, m}\right}$ – meiji163 Jun 20 '15 at 04:20
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Let $\,\ d = (M,X),\,\ m = M/d,\,\ x = X/d.\ $ Then
$\quad M\mid X(2^k\!-1)\!\!\overset{\ \div\,d}\iff m\mid x(2^k\!-1)\overset{(m,x)\,=\,1}\iff m\mid 2^k\!-1\iff {\rm ord}_m(2)\mid k$
Bill Dubuque
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My reading abilities are limited so this answer does not really make much sense to me. – Remy Hartfield Jun 20 '15 at 04:39
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I uppercased m and x, and converted the congruence to divisibility form, then cancelled their gcd $,d,,$ then applied Euclid's Lemma. – Bill Dubuque Jun 20 '15 at 04:50
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@Remy If any steps remain unclear then let me know precisely where, so I can elaborate. – Bill Dubuque Jun 20 '15 at 05:04