$(x,y) \in \mathbb Z \times\mathbb Z$ with $336x+378y=\gcd(336,378)$ $(x,y) \in \mathbb Z \times\mathbb Z$ with $336x+378y=\gcd(336,378)$
Question: How can I get every possible combination of $x$ and $y$?
My solution so far: 
First I have calculated the $\gcd(336,378)=42$.
So using that I have $42= 1\cdot378-1\cdot336$
So $x=1$ and $y=-1$.
To get the general formula I have tried two things with gcd and lcm as a factor:
gcd:
$42=1 \cdot 42 \cdot x \cdot 378- 1 \cdot 42 \cdot x \cdot 336, x \in \mathbb Z$
lcm:
$42=1 \cdot 3024 \cdot x \cdot378 - 1 \cdot 3024 \cdot x, x\in \mathbb Z$
Unfortunately both do not make sense - it was just a guess. 

You have found the minimum solution to Bezout's identity where $x=-1,y=1$
Now all the solutions are given in pairs by:
$x+k\frac{b}{\gcd(a,b)},y-k\frac{a}{\gcd(a,b)}$
For example, 
$k=1 \rightarrow x=8,y=-7$ and $8\cdot 336 + (-7)\cdot 378 = 42$
$k=2 \rightarrow x=17,y=-15$ and $17\cdot 336 + (-15)\cdot 378 = 42$
