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I'm thinking about

$ax \equiv 0 \mod m$ $\Leftrightarrow$ $m \mid (ax - 0) \Leftrightarrow m\mid ax$

Since $\gcd(a,m) \not= 1$, we have

$m \mid a \Rightarrow m\mid ax$

But I'm not tottaly sure about this. Any hints?

Michael Hardy
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dumb_undergrad
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    If $d=\gcd(a.m)$ then $x=\frac md$ is a solution. – Hagen von Eitzen Jun 12 '17 at 16:00
  • No, if the gcd of $a$ and $m$ is not 1, then you cannot infer that $m$ divides $a$. – Wuestenfux Jun 12 '17 at 16:01
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    That last line is trivial and not broad enough. $m|a \implies m|ax$ whether or not $\gcd(m,a) = 1$ and there is utterly no reason to assume $m|a$. Indeed if $m \ne\gcd(m,a)$ then $m \not \mid a$. My hint: Just do it. if $\gcd(a,m) = d$ then $a = a'd; m = m'd$ $ax = km \implies a'x = km'$ Let $k=a'$ and $x = m'=\frac m{\gcd(a,m)}$ will be a solution. As 1 < m' < m; m' != 0. – fleablood Jun 12 '17 at 16:39
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    You can find more conceptual ways to view the proof in the linked duplicate. Assuming $a\not\equiv 0,$ then $ax\equiv 0,$ has a solution $x\not\equiv 0\iff x$ is a zero-divisor (by definition). So you question si to show that $a,m$ not coprime $\Rightarrow x$ is a zero-divisor. – Bill Dubuque Jun 12 '17 at 20:51

3 Answers3

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Let $d=\gcd(a,m)$, $m/d$ is a solution since $a(m/d)=(a/d)m$ and $a/d$ is an integer.

Michael Hardy
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Tsemo Aristide
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Let $d$ be the gcd of $a$ and $m$. Then $$ m\mid ax\implies m/d\mid|(a/d)x $$ But $m/d$ and $a/d$ are coprime. So what can we conclude?

Sri-Amirthan Theivendran
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Just do it:

$ax \equiv 0 \mod m$ means there exist some integer $k$ so that $ax = km$. There are an infinite number of such $k$ because $ax = km\implies x = \frac {km}a$ and we can alway just choose $k$ to be a multiple of $a$.

Those would be trivial solutions. If $a|k$ then let $x = \frac ka $ and $ax = km$ is a multiple of $m$ and $x \equiv 0 \mod m$.

So this is a matter of finding $x = \frac {km}a\in \mathbb Z$ and $a\not \mid k$.

This means $m$ and $a$ have a non-trivial factor in common. Which... they do, of course. Let it be $d =\gcd(a,m) \ne 1$. Then

$x = \frac {km} {a} = \frac {k*\gcd(a,m) * (\frac m{\gcd(a,m)})}{\gcd(a,m)*\frac a{\gcd(a,m)}}=\frac {k*\frac m{\gcd(a,m)}}{\frac a{\gcd(a,m)}}= \frac{k}{\frac a{\gcd(a,m)}}\frac m{\gcd(a,m)}$

$k$ can be any multiple of $\frac {a}{\gcd(a,m)}$

....

In other words.

SOLUTION:

Let $d = \gcd(a,m)$. Let $a = a'd; m = m'd$. Then if $x = m'$:

$ax = am' = a'm'd = a'm \equiv 0 \mod m$.

If $d \ne 1$ then $1 < m' < m$ and $x = m' \not \equiv 0 \mod m$.

fleablood
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