Does exists any better way to solve "Simple hit problem" ? $y = \frac{a x + b}{c}$

Question

I am looking for other solution of a problem, which I named:
$$\text{Simple hit problem}$$

Let:
$$y = \frac{a x + b}{c}$$ $$a, b, c, x \in \Bbb{Z}$$ For which $a, b, c$ there exists $x$ for such $y \in \Bbb{Z}$ ?

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I noticed that the solution exists for $a, b, c$, if:
$$(ax) \ \text{mod} \ c = c - (b \ \text{mod} \ c),$$ but the approach relies on checking all unique values of result of modulo operator from left side of equation.

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The question is, if there exists any better approach for this problem or anyway it come down to check, if some equation will be true for some $x \in A$, where A will be a set of finite ammount of values ?

Answer 1

We can rewrite the equation as $ax+b=yc$, or by renaming $a$ with $-a$ to the Diophantine equation $ax+cy=b$. This has been solved here and is a standard question:

Diophantine equation $ax + by = c$ has an integer solution $x_0, y_0$ if and only if $\gcd(a,b)|c$

Answer 2

Let:
$$y = \frac{a x + b}{c}$$ $$a, b, c, x \in \Bbb{Z}$$ For which $a, b, c$ there exists $x$ for such $y \in \Bbb{Z}$ ?

Question seems ambiguous to me. As other responses have done, a reasonable assumption is that $x$ must be an integer.

However, the other responses seemed to have interpreted the equation as that $x$ and $y$ may vary. The alternative interpretation is that you are given some fixed integer $y$, and are asking for which values of $a,b,c$ can a solution involving an integer value of $x$ be produced.

Under this alternative interpretation of the problem, which (admittedly) the OP may not have intended, since

$cy = ax + b$,
the constraints on $a,b,c$ for the fixed value of $y$ are that
$c \neq 0$ and
$a|(cy - b)$.

Here, if $b,c$ are chosen so that $(cy - b) = 0$, then $a$ must equal $0$, and $x$ can be anything.

Alternatively, if $b,c$ are chosen so that $(cy - b) \neq 0$, then $a$ can not equal $0$.