How can I solve $23^{2m}\equiv 73 \left (\text{mod } 1971 \right )$

Question

Solutions for $m \in{N} $. The original equation says that $27 \times 23^{2m}\equiv 0 \left ( \text{mod }1971 \right), $ and therefore I concluded that the congruence above holds, because $27\times 73 = 1971$. I hope my conclusion is true and that you can help me.

Answer 1

The original equation says that $27 \times 23^{2m}\equiv 0 \left ( \text{mod }1971 \right) $

From this, you should have concluded that $23^{2m}\equiv0\bmod73$.

($27\times 23^{2m}\equiv0\bmod1971\iff 1971|27\times23^{2m}\iff73|23^{2m}\iff23^{2m}\equiv0\bmod73.)$

Furthermore, this has no solutions, because $23$ and $73$ are different primes, so $73\nmid23^{2m}$.