<s>[INST] Find a $k$ such that $3^k \equiv -6 \pmod{43}$ I have been trying to find this $k$, but I am stuck.
The only information I could extract was from the Fermat's Little Theorem:
Since $43$ doesn't divide 3 and it is a prime, it follows $3^{42} \equiv 1 \pmod{43}$
However, I have no idea how to proceed from now.
All help is appreciated!
 [/INST] Notice $3$ and $43$ are relatively prime.  Thos if we hae $3m \equiv 3n \pmod{43}$ we can safely conclude that $m \equiv n\pmod{43}$
So if $3^k \equiv -6 \pmod {43}$ then 
$3^{k-1} \equiv -2\equiv -45\pmod {43}$
$3^{k-2} \equiv -15$
$3^{k-3}\equiv -5\equiv -48$
$3^{k-4} \equiv -16\equiv 27$
$3^{k-7} \equiv 1\pmod {43}$.
So we can let $k = 7$.
Another way of looking add it is that as $43 \equiv 1 \pmod 3$ we can find a multiple of $3$ by adding or subtraction $43$.
$-6 \equiv 3(-2)\equiv$
$3(-45)\equiv 3^3(-5)\equiv 3^3(-48)$
$3^4(-16)\equiv 3^4(27)\equiv 3^7$.
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