Find any two integers that satisfy $19x+47y = 1$ How can I find two integers that satisfy $19x+47y = 1$? Is there some technique to finding 2 numbers? I can't find any 2 numbers in $\mathbb{Z}$ that make this work. 
Thank you!

Consider the continued fraction of $\frac{47}{19}$:
$$ \frac{47}{19}=2+\frac{1}{\frac{19}{9}}=2+\frac{1}{2+\frac{1}{9}}=[2;2,9] $$
truncate it and expand it back:
$$ [2;2] = 2+\frac{1}{2} = \frac{5}{2}. $$
We have that $\frac{5}{2}$ and $\frac{47}{19}$ are consecutive convergents of the same continued fraction, hence their difference is $\pm\frac{1}{2\cdot 19}$.
Indeed 
$$ \frac{47}{19}-\frac{5}{2}=\frac{94-95}{38}=\frac{-1}{38} $$
leads to
$$ 47\cdot 2 - 5\cdot 19 = -1 $$
$$ 47\cdot(-2) + 19\cdot 5 = 1 $$
$$ 47(19-2)+19(5-47) = 1 $$
$$ \color{red}{47\cdot 17 - 19\cdot 42 = 1.}$$
