TITLE: Solution to exponential congruence
QUESTION [6 upvotes]: Is there a clever solution to the congruence without going through all the values of x up to 58?$$2^x \equiv 43\pmod{59}$$
Can I somehow use the fact that  $2^4 \equiv -43\pmod{59}$ ?

REPLY [2 votes]: Somebody posted a very nice solution using quadratic residues, but here is an easier, and slightly more "dirty" solution. 
Note that $$2^{29} \equiv (2^6)^{4} \times 32 \equiv 5^4 \times 32  \equiv 35 \times 32 \equiv 7 \times 160 \equiv 7 \times 42 \equiv 294 \equiv -1 \pmod {58}$$
Also note $$2^2 \equiv 4 \pmod {58}$$
This implies that $2$ is a primitive root. Since $$2^{33} \equiv 2^{29} \times 2^{4} \equiv -16 \equiv 43 \pmod {58}$$
We have that $33+58k$ is the only possible solution where $k$ is an integer.