Result $a \equiv b \mod m$ or $c \equiv d \mod m$ is false $\Rightarrow ac \equiv cd \mod m$ is false? Root extraction $\mod m$ I know that $a \equiv b \pmod{m}$ and $c \equiv d \pmod{m}$ implies $ac \equiv bd \pmod{m}$.
However, can one show that if $a \equiv b \pmod{m}$ is false, then:
$ac \equiv bd \pmod{m}$ is false, when $c \equiv d \pmod{m}$ is false.
and
$ac \equiv bd \pmod{m}$ is false, when $c \equiv d \pmod{m}$ is true.
That is the product of two congruences modulo $m$ is always false, if one of them is false ?
I need this result to understand how to extract roots modulo $m$.
$x^k \equiv b \pmod{m}$ has the solution $b^u$, since one can find $u, w$ such that:
$x^k \equiv b \Rightarrow b^u \equiv (x^k)^u \equiv x^{1+\theta(m)w}\equiv x \pmod{m}$ where $\theta$ denote Euler's Totient function.
Here I see that $x$ is congruent to $b^u$ and $(x^k)^u$ but how to I do tell if $b^u \equiv (x^k)^u \Rightarrow x^k \equiv b \pmod{m}$ ?
Thanks for your time.

Isn't this as simple as noting that $1\cdot k \equiv k\cdot 1\pmod m$ for all $k$? So if we take $k$ such that $k\not\equiv 1\pmod m$ then we have a counter-example to your claim.
