problem about modulus root and quadratic reciprocity

Question

How to calculate $x$ from $x^{14} \equiv 26 \pmod{91}$?

What I tried: Let $y=x^2$ $$y^7 \equiv 26 \mod 91$$ then $y \equiv 26 \mod 91$.

Then I have $x^2 \equiv 26 \mod 91$ How to solve this? or this cannot be solved?

I have tried to use the quadratic reciprocity: $$\frac{26}{91} = \frac{26}{7}\left(\frac{26}{13}\right)$$

However $\frac{26}{13}=0$, I don't know what to do with this. Thanks.

Answer 1

Use the Chinese remainder theorem. Your $x^{14} = 0 \mod 13,$ therefore so is $x.$ And $x^{14}$ is $5 \mod 7,$ but recall that $x^7 = x \mod 7.$

Answer 2

Hint $\rm\ \ mod\ 7\!:\ {-}2\equiv x^{14}\, \overset{cubing}\Rightarrow \color{#c00}{-1}\equiv (x^6)^7\overset{Fermat}\equiv \color{#c00}1\ \Rightarrow\Leftarrow$

Key Idea $\ $ Nonzero squares (like $\rm\,x^{14})$ are cube-roots of $1,$ but $-2$ is not.