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.
 A: An alternative to the example of André Nicolas is to consider the case $a \not\equiv b \operatorname{mod} m$ and $0 \equiv 0 \operatorname{mod} m$.
In general, your last statement is false. That is, if $a^n \equiv b^n \operatorname{mod} m$, then it is possible that $a \not\equiv b \operatorname{mod} m$. For example, $1^2 \equiv 1 \operatorname{mod} m$ and $(m-1)^2 \equiv 1 \operatorname{mod} m$, but for $m > 2$,  $1 \not\equiv m-1 \operatorname{mod} m$.
A: We cannot. Note that $1\equiv 2\pmod{5}$ is false, as is $7\equiv 6\pmod{5}$, but $1\cdot 7\equiv 2\cdot 6$ is true. 
(For the current question, which presumably has a typo, one can use $c\equiv 0\pmod{5}$.)
For extracting roots modulo $m$, there cannot be a full discussion unless we know more about $b$, $m$, and $k$. In many cases there are no solutions. 
Remark: Suppose that $b$ and $m$ are relatively prime, and $k$ and $\varphi(m)$ are relatively prime. Then we can use Euler's Theorem to solve $x^k\equiv b\pmod{m}$.
For there exist integers $s$ and $t$ such that $sk+t\varphi(m)=1$. Then 
$$b^1=b^{sk+t\varphi(m)}=b^{t\varphi(m)}(b^s)^k\equiv (b^s)^k\pmod{m}.$$
A: 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.
