$( x \cdot y ) \mod 37 = 1$ I am doing a paper for my security class. I have this equation which I'm trying to understand
                                  $$( x \cdot  y ) \mod 37 = 1 $$
e.g.  if  $x = 8$ and $y = ?$ ; which has to be the inversion of x . then y in this case is 14. 
My question is how do I solve an equation like this? and is it possible that value of $y$ is a negative number? if so, then an example would be great.
Thank you in advance !

Using this or this , $2$ is  a primitive root of $37\iff $ord$_{37}2=\phi(37)=36$
So, $2$ is a generator of this cyclic group $\mod {37}$
Now as, $\displaystyle2^{36}\equiv1\pmod{37}, 2^{-a}\equiv2^{37-a}\pmod{37}$
As, $\displaystyle8=2^3, 8^{-1}\equiv 2^{-3}\equiv2^{36-3}\pmod{37}$
Now, $\displaystyle2^5=32\equiv-5\pmod{37},2^6\equiv-10,2^7\equiv-20,2^8\equiv-40\equiv-3,2^{32}\equiv(-3)^4\equiv7$
$\displaystyle\implies 2^{33}\equiv2\cdot7\pmod{37}$
