The function is: $y = \frac{58-3x}{2x+1}$
where y and x are positive integers. Is there a way to simply tell that there is a solution for this equation (other than x=0)?
I see that x cannot be even and I made a spreadsheet to get the solution by trying but I couldn't find a more simple way.
If $d$ divides both $58-3x,2x+1$
$d$ must divide $3(2x+1)+2(58-3x)=119$
So, $2x+1$ must divide $119$ to keep $y$ an integer
Check if $2x+1$ one of $\{\pm1,\pm7,\pm17,\pm119\}$
As $x\ge1,2x+1\ge3,2x+1$ can be one of $\{7,17,119\}$
For example, if $2x+1=7, x=3,58-3x=49$
Alternatively, $$2y+3=\dfrac{116-6x}{2x+1}=\dfrac{119}{2x+1}$$
Now $\dfrac{119}{2x+1}=2y+3\ge5,2x+1\le\dfrac{119}5=23.8<24$