Find all solutions $x^{40} \equiv 2 \pmod {79}$

Question

Find all solutions $x^{40} \equiv 2 \pmod {79}$.

I tried following steps from this question.

So we get:

$$y^{40} \equiv 2 \pmod {79} \implies (y^{40})^2=y^{80} \equiv 2 \pmod {79}.$$

Now since $80$ gives remainder $1$ when divided by $\!\!\mod {79}$, then: $$y^{80} \equiv y^{79} \cdot y\equiv y \pmod {79}.$$ Thus, $x=2$? I did a direct computational of $2^{40} \equiv 2 \pmod {79}$ and it does seem true.

Did I approach this correctly?

Answer 1

Your basic idea is sound, but there are a number of arithmetic errors in your calculation. A corrected version would be something like:

$$x^{40}\equiv 2 \implies x^{80}\equiv 4\implies x^2\equiv 4\implies x\equiv \pm 2$$

Where every congruence is $\pmod {79}$.