Vtyjkyuk

There are two similar congruences:
$$
x^2-6xequiv16pmod512\
y^2-yequiv16pmod512
$$
It is easy to see that the $gcd$ of all three parts in both of them is $16$, $x$ and $x-6$ are even and one of $y$ and $y-1$ is odd. After also noticing $xnotequiv x-6pmod4$ we have
$$
x equiv 0 pmod8\
or\
x-6 equiv 0 pmod8
$$
$$
y equiv 0 pmod16\
or\
y-1 equiv 0 pmod16
$$
Making substitutions $x=8n,x=8n+6,y=16m,y=16m+1$ I obtained congruences with modulus $32$, but couldn't make it any further. I also discovered that the first congruence can be rewritten as $(x+2)(x-8) equiv 0pmod512$ while $-2$ and $8$ appear to be its solutions. However, it's still not clear to me whether there are more of them.

What am I missing? How can such congruences be solved?

The first one is easy. You need $(x-8)(x+2)$ divisible by $512$. Both $x+8$ and $x-2$ are even, but only one of them can have higher powers of $2$ as factors. Thus, $xequiv 8pmod 256$ or $xequiv -2pmod256$.

The second requires more care.

First assuming $yequiv 0pmod16$ you have $y=16y_0$ and $$16y_0^2=y_0+1pmod32$$

Since $16y_0^2$ is even, $y_0+1$ is even, so $y_0$ is odd. When $y_0$ odd, $16y_0^2equiv 16pmod32$, so $y_0equiv 15pmod 32=$ and $yequiv 16cdot 15=240pmod512$.

Second, assume $y=16y_0+1$. Then $$16y_0^2 equiv 1-y_0pmod32$$ So $y_0$ is odd, $1-y_0equiv 16pmod 32$, or $y_0equiv 17pmod32$. So $yequiv 16cdot 17+1=273pmod512$

So the two solutions are $yequiv 240,273pmod512$.

Note that the sum of these two values is $1pmod512$, as befits the two roots of any quadratic equation of the form $x^2-x+C=0$.

Indeed, we have $(y-240)(y-273)equiv y^2-y-16pmod512$, so these are the only solutions, since only one of $y-240$ and $y-273$ can be even.