I'm trying to solve a form of modular arithmetic I've never seen before. I'm completely stuck. Any hints in how to crack this would be of great help.
$$ -18 \equiv 19y \pmod{1967-y}$$
Or similarly, how do I find integer solutions for:
$$ y = \frac{-18-1967k}{19-k}$$
On
Hint $\,\ 1967\!-\!y\mid 19y\!+\!18\iff (1967\!-\!y)x = 19y\!+\!18\, $ for some integer $\,x.\, $ In standard form
$$ xy -1967 x + 19 y\, = -18\ \ \ $$
Completing a square generalizes to completing a product above, namely
$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff &&\!\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$
Your $\,ad\!+\!bc\,$ is a product of two primes $\,\Rightarrow\,$ few possibilities for its factors $\,ax\!+\!c,\ ay\!+\!b.$
Since there exists an integer $k$ such that $$-18-19y=k(1967-y),$$ we have $$y=\frac{1967k+18}{k-19}=1967+\frac{18+19\times 1967}{k-19}=1967+\frac{139\times 269}{k-19}$$$$\iff 1967-y=\frac{139\times 269}{19-k}\gt 0$$ where $139$ and $269$ are primes.
Since $19-k\gt 0$ has to be a positive divisor of $139\times 269$, we have $$19-k=1,\ 139,\ 269,\ 139\times 269\Rightarrow 1967-y=139\times 269,\ 269,\ 139,\ 1$$ $$\Rightarrow y=1967-139\times 269,\ 1967-269,\ 1967-139,\ 1967-1$$ $$\Rightarrow y=-35424,\ 1698,\ 1828,\ 1966.$$