Solve $x^{83}\equiv 7\pmod {139}$

Let $g$ be a generator of $\mathbb{Z}/139\mathbb{Z}$, that is, a primitive element. Since $g$ is a generator, there exists $n$ such that $g^n = 7$. Now, by little Fermat's theorem, $g^{p-1} \equiv 1 \pmod p$, so $g^{138} \equiv 1 \pmod{139}$, so $g^{n+138k} \equiv g^n \equiv 7 \pmod{139}$.

It is now enough to find $k$ such that $138k+n = 83m$ for some $m$. At this point, $g^{n+138k} = g^{83m}$, and so we can pick $x = g^m$, so that $x^{83} = (g^m)^{83} = g^{83m} = g^{n+138k} \equiv 7 \pmod{139}$ .

Now, I don't think there is any efficient method for finding a generator other than bruteforcing it. In our case, $139-1 = 138 = 2 \cdot 3 \cdot 23$, so to check if $g$ is a generator it is enough to check if $g^{138/2}$, $g^{138/3}$ and $g^{138/23}$ are all different from $1$. If we try $g = 2$, we have $2^{69} \equiv -1 \pmod{139}$, $2^{46} \equiv 96 \pmod{139}$, and $2^6 = 64 \pmod{139}$, so none of them is $1$ and so $2$ is a generator.

Now we need to find $n$ such that $2^n \equiv 7 \pmod{139}$. Again, I think the most efficient way is to bruteforce it; we find $n=50$.

Finally, we need $k$ such that $138k+50$ is a multiple of $83$. Now this is easier: $138 \equiv -28 \pmod{83}$, and it is easy to see that $28 \cdot 3 = 84 \equiv 1 \pmod{83}$, so the multiplicative inverse of $-28$ in $(\mathbb{Z}/83\mathbb{Z})^*$ is $-3$. Now, we wanted to solve $138k+50 \equiv 0 \pmod{83}$, so $50 \equiv -138k \pmod{83}$, so $3 \cdot 50 \equiv 3 \cdot (-138k)$, so $150 \equiv k$ (because $3 \cdot (-138) \equiv 3 \cdot 28 = 84 \equiv 1$), so $k = 67$.

Putting everything together, we need $m$ such that $83m = 67 \cdot 138 + 50$, so $m=112$. It follows that $(2^{112})^{83} \equiv 7 \pmod{139}$, as desired. To simplify it slightly, one can check that $2^{112} \equiv -12 \pmod{139}$, so $(-12)^{83} \equiv 7 \pmod{139}$.