Prove that the modular congruence holds: $b^d$ $=$ $r \pmod n$, $b^{d/q}$ $=$ $x \pmod n$, then $x^q$ $=$ $r \pmod n$.

Question

Prove that if

$b^d$ $=$ $r \pmod n$

$b^{d/q}$ $=$ $x \pmod n$,

then

$x^q$ $=$ $r \pmod n$

for any integers $b$, $n$, $r$, and $q$ (which divides $d$).

Or more simply that $b^d$ $=$ $x^q$ $\pmod n$ if the first two conditions above are true.

Answer 1

If $x\equiv b^{d/q} \mod n$, then $x^q \equiv (b^{d/q})^q=b^d \mod n$.

Answer 2

By the Congruence Power Rule we can raise the $2$nd congruence to power $\,q,\,$ i.e.

$$\begin{align} (x\, &\equiv\, b^{\large \,d/q})^{\large \,q}\\[0.1em] \Rightarrow\,\ x^{\large q} &\equiv\, b^{\large d} \equiv\, r \end{align}\quad $$