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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.


marked as duplicate by Arnaud Mortier, gt6989b, José Carlos Santos, Parcly Taxel, user99914 Jul 27 '18 at 19:22

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  • $\begingroup$ The linear equations determine two straight lines of different slopes that must intersect in exactly one point. Solve for that point and see if it just happens to have integer components. $\endgroup$ – David G. Stork Jul 24 '18 at 12:07
  • $\begingroup$ No sorry they are not simultaneous. $\endgroup$ – Roskiller Jul 24 '18 at 12:08
  • 1
    $\begingroup$ If they are not simultaneous equations you work on each one separately with the extended Euclidean algorithm en.wikipedia.org/wiki/Extended_Euclidean_algorithm (the way the powers of $2$ and $3$ determine the coefficients says that there will be solutions. $\endgroup$ – Ethan Bolker Jul 24 '18 at 12:09
  • $\begingroup$ @DavidG.Stork you misread the post, that's two separate Diophantine equations. $\endgroup$ – Arnaud Mortier Jul 24 '18 at 12:09
  • $\begingroup$ Try solving it @roskiller, then verify if the solutions are whole numbers or not. $\endgroup$ – Ahmad Bazzi Jul 24 '18 at 12:09

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.

By the way,




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.


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