How to find integer solutions of $x^2 + by = c$

Question

I can find lots of examples for $ax + by = c$, but how do I get started with the $\text{^2}$ there?

My specific equation is; $2x^2 - 154y = 116$

I thought I would reduce and substitute to get a familiar form; $s - 77t = 58$

which I realized has a trivial solution; $(s_0,t_0) = (58,0)$

I'm not particularly sure where to go from here, or even if this is the right path to take.

Answer 1

Suppose $x$ and $y$ are integers such that $$2x^2-154y=116.$$ As you note, we can divide everything by $2$ to get $$x^2-77y=58.$$ Reducing mod $7$ and $11$ then shows that $$x^2\equiv2\pmod{7} \qquad\text{ and }\qquad x^2\equiv3\pmod{11},$$ respectively, and hence $x\equiv3,4\pmod{7}$ and $x\equiv5,6\pmod{11}$. Then by the Chinese remainder theorem $$x\equiv17,38,39,60\pmod{77}.$$ On the other hand, for any integer $x$ satisfying the congruence above we can define the integer $y$ as $$y:=\frac{2x^2-116}{154}=\frac{x^2-58}{77},$$ which is an integer precisely because $x^2\equiv58\pmod{77}$, and so we have $$2x^2-154y=116.$$

Answer 2

I have an inelegant solution to this problem.

Consider your rearrangement: $$s^2 - 77t = 58$$ $$s^2 \equiv 58~\texttt{mod}~77$$ I imagine there is a more elegant way to figure this out, but my approach was to consider integers 0 through 77 and determine which squared to 58. These are your $s$ modulo 77, from there you should be able to figure out your $t$.

Edit to add: You can limit your search a bit since you know that 58 is relatively prime from 77 so you don't need to check anything that is a multiple of 11 or 7.