Do there exist whole number solutions to $27y + 23 = 32x$ and $81y + 85 = 128x$?

Question

So I think I found these $$27y + 23 = 32x$$ $$81y + 85 = 128x$$ in a text-book or something, and it was a graphing problem. (These are not simultaneous equations, they are separate.)

I tried to find integer solutions to this and after putting in some numbers I still couldn't find any.

I used some graphing software and still could not find any integer solutions for $x, y \in \Bbb Z$.

So I wonder, do any solutions exist? But more importantly, is there a technique for checking if equations like these actually have integer solutions, if so what is the technique.

This isn't overly important, but if there was such a technique that would be helpful.

Thank you.

Answer 1

By the Bézout identity, these two equations do have solutions.

The criterion for existence is that the $\gcd$ of the coefficients of $x$ and $y$ must divide the constant term.

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By the way,

$$27\cdot11+23=32\cdot10,$$

$$81\cdot59+85=128\cdot38.$$

Answer 2

You may write $x={27y+23\over 32}$ and observe whether $27y+23$ is divisible by $32$.
Now, $x=y+1-({5y+9\over 32})$ which gives $5y+9$ should be divisible by $32$. Now observe that $5y$ ends with $5$ or $0$. That means $5y+9$ either is of form $10k+4$ or $10k+9$. We know that some multiples of $32$ end with $4$. Hence there are whole number solutions.