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Prove that the linear congruence mx = p mod q, where gcd of (m,q) = 1, has a unique solution x = $p_0$ mod q, where $p_0$ belongs to {(p + qk)/m} where k = 0 to m-1.

Also, prove that, If gcd of (m, q) = d and d|p, then the linear congruence mx = p mod q, has d distinct (in-congruent) solutions modulo q.

EDIT also Prove that, the linear congruence cx = a mod b, where gcd(c, b) = 1 has solution $x = a(1-b^{\phi(c)})/c \mod b$.

EDIT The third question of my previous post can be read as follows. The linear congruence $cx \equiv a \pmod b$ and $\gcd (c, b) = 1$ has solution in the form of $x \equiv a(1-b^{\phi(c)})/c \pmod b$.

A standard method of solving linear congruences involves Euler's phi function [2,3], or totient, denoted by $\phi$. The totient $\phi(b)$ enumerates the positive integers less than $b$ which are relatively prime to $b$. Euler's extension of Fermat's theorem states that $c^{\phi(b)} \equiv 1 \pmod b$, if $\gcd(c, b) = 1$ and multiplying the linear congruence $cx \equiv a \pmod b$ through by the factor $c^{\phi(b)-1}$ gives $x \equiv ac^{\phi(b)-1} \pmod b$.

Edited If gcd(c, b) = d and d|a, then the linear congruence cx = a mod b has d distinct solutions x = [$x_0$, $x_0$ + $b_0$, ... , $x_0$ + $b_0$(d - 1)] mod b, where a = $a_0$ d, b = $b_0$ d, c = $c_0$ d, and $x_0$$ \equiv $$a_0$$(1-$$b_0$$^{\phi(c_0)})/$$c_0$$ \pmod b$. Can you generalize?

Gandhi
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    See this answer. – Arturo Magidin Aug 18 '11 at 20:37
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    @Gandhi, the first two parts are standard results about linear congruences, discussed in many intro Number Theory textbooks. The proofs are a bit long to write out, if one starts by not assuming any previous knowledge, so it's probably best if you seek out a textbook, or possibly search the web for "linear congruence" or some such keyphrase. The third question, all these symbols $a$, $b$, $c$, and $k$ seem to come from nowhere, with no explanation. Maybe if you gave an example it would be clearer just what you are talking about. – Gerry Myerson Aug 19 '11 at 13:48
  • Thank you for all posts of comments. I am giving little more explanation for third question of my post. Please see my edited one. – Gandhi Aug 20 '11 at 15:06
  • @Arturo and others involved in the previous discussion: In view of Gandhi's most recent edit, I am removing the long (off-topic) discussion on etiquette. @ Gandhi: still, it would be better if you would edit the original post to read "How can I prove such and such fact?" instead of "Prove such and such fact." On a Question and Answer site, we prefer it if you ask questions, and not issue demands. – Willie Wong Aug 20 '11 at 16:02
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    I have edited some TeX into part of the question to improve readability. There is something very strange in the third sentence, which begins with $m,p,q$ but ends with $a,b,c$. I hope OP will edit to clarify. I also left in the [2,3] in the last part though I don't know what it means. The "third question" doesn't seem to be a question, as the last paragraph seems to answer it. Gandhi, is there still something you want to know? – Gerry Myerson Aug 20 '11 at 23:46
  • Thank you for editing and leave the term [2,3] and answer for the rest – Gandhi Aug 21 '11 at 08:22

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We're asked to prove that if $\gcd(b,c)=1$ then the congruence $cx\equiv a\pmod b$ has the solution $x\equiv a(1-b^{\phi(c)})/c\pmod b$, where $\phi$ is the euler phi-function.

Note that by Euler's Theorem $b^{\phi(c)}\equiv1\pmod c$ so $(1-b^{\phi(c)})/c$ is an integer. If $x\equiv a(1-b^{\phi(c)})/c\pmod b$ then $cx\equiv a(1-b^{\phi(c)})\equiv a\pmod b$ since obviously $b^{\phi(c)}\equiv0\pmod b$. So we have proved that the formula for $x$ gives a solution to the congruence.

Gerry Myerson
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