<s>[INST] Finding GCD of $95$ and $39$ My Algebra instructor gave us this problem which is to find gcd of $95$ and $39$ and express it as $95x+39y$.  Also are $x$, $y$ unique?
Now they are relatively prime so GCD is $1$.  I have no clue how to find $x$ and $y$ and if they are unique or not.
Thanks!
 [/INST] Although you could use the extended Euclidean algorithm to calculate that value, sometimes it's simpler to use Euler's theorem and exponentiation by squaring.
I don't want to solve your problem for you, so I will use similar numbers $94$ and $51$ instead.
More precisely, say we want to find $x$ and $y$ such that $94x+51y = \gcd(94,51) = 1$. This implies that $94x=1 \pmod{51}$, thus 
\begin{align}
x & = 94^{\phi(51)-1}    &\pmod{51} \\
  & = 94^{(17-1)(3-1)-1} &\pmod{51} \\
  & = 94^{31}            &\pmod{51} \\
  & = 94^{30+1}    &\pmod{51} \\
  & = 94\cdot(94^{15})^2 &\pmod{51} \\
  & = 94\cdot(94\cdot(94\cdot(94\cdot (94\cdot1^2)^2)^2)^2)^2 &\pmod{51} \\
  & = 94\cdot(94\cdot(94\cdot(94\cdot (94\cdot1^2)^2)^2)^2)^2 &\pmod{51} \\
  & = 19 &\pmod{51}
\end{align}
That means we can pick any $x = 19 \pmod{51}$ to get a solution. For example when $x = 19$ then $94\cdot 19 + 51y = 1$ gets us $y = \frac{1-94\cdot 19}{51} = -35$. On the other hand if $x = 19+51 = 70$ then $y=-129$.
I hope this helps $\ddot\smile$
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