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I'm quite new to modular arithmetic. Therefore, this question may be very silly. Sorry for this.

Suppose that I want to prove that: $$a \cdot b \equiv 0 \mod{c}.$$

If I know that

$$c \equiv 0 \mod{a},$$

then can I conclude that:

$$b \equiv 0 \mod{d} \Rightarrow a \cdot b \equiv 0 \mod{c},$$

where $c = a \cdot d$?

Bill Dubuque
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the_candyman
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    If you know $c \equiv 0 (mod $ $a)$, you can write $c=k.a$ for some integer $k$. Now you want to prove that $a.b=m.c$ for some integer $m$ Now substitute $c=k.a$ So, $a.b=m.a.k$ and thus $b=m.k$ . So yes, what you said is correct. – Aditya Mar 22 '20 at 11:48
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    You don't need $ab\equiv0\bmod c$. From $b\equiv0\bmod d$, you just multiply through by $a$ to get $ab\equiv0\bmod{ad}$, but $ad=c$. – Gerry Myerson Mar 22 '20 at 12:06

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Yes, it's simply $\ \bbox[5px,border:1px solid #c00]{ad\mid ab\iff d\mid b}\,\ $ by cancelling or scaling by $\,a\neq 0,\ $ i.e.

$\ \, \dfrac{ab}{ad} = \dfrac{b}d,\ $ so $\,\ ad\mid ab\iff \dfrac{ab}{ad}\in \Bbb Z\iff \dfrac{b}{d}\in\Bbb Z\iff d\mid b$

Alternatively, eliminating fractions and emphasizing an equational view

$$ ad\mid ab\iff \exists x\!:\, ad\,x = ab \iff \exists x\!:\, d\,x = b\iff d\mid b $$

i.e. equationally: $\ \,ad\,x = ab\,$ has an integer root $x\!\iff\! d\,x = b\, $ has an integer root $x$

with very easy proof: cancelling $\,a\neq 0\,$ yields $(\Rightarrow)$, and scaling by $\,a\,$ yields $(\Leftarrow)$.

i.e. the operation of "scaling by $a$" is invertible if $a$ is cancellable, so scaling by $a$ and cancelling $a$ yield equivalent equations (same set of roots), so these operations preserve solvability.

This preservation of equations is well known in the special case that $a$ is invertible (e.g. $\,a\neq 0\,$ in a field like $\,\Bbb Q,\Bbb R,\Bbb C)$ when we normalize polynomials to be monic (lead coef $=1$) by scaling by $a^{-1}$.

Alternatively it can be viewed as a special case of congruence scaling / division / cancellation when written in the language of congruences (vs. divisibility).

Bill Dubuque
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