Question: Find an integer that solves the congruence $$x^{83}\equiv 7 \pmod{139}$$
My working: Let $b$ denote some primitive root of the prime mod 139, and let $$x\equiv b^y\mod 139$$ for some integer $y$. Likewise, let $$7\equiv b^k \mod 139$$ for some integer k. Then, $$b^{83y}\equiv b^k \mod 139$$, and by a property of primitive roots, $$83y\equiv k \mod 138$$. So after solving for $y$, and then going back to find $x$, we have $x\equiv b^y \equiv b^{83^{-1}k}\equiv(\underbrace{b^k}_{7})^{83^{-1}}\equiv 7^{83^{-1}} \mod 139$. Then from using the computer calculator (or otherwise), I got $83^{-1}\equiv67 \mod 139$. So my final answer is $x\equiv122 \mod 139$, but when I check with the computer, $122^{83}\not\equiv7\pmod{139}$ (in fact, the solution seems to be 127), which confuses me. Did I make a mistake / do a fallacious step somewhere?
