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How do you solve a Diophantine equation of the kind ax + by = c or ax + by + cz = d when b is divisible by a?

I understand how to solve a Diophantine equation in general (ex. 132x + 52 y = 6), but I'm having trouble applying Bézout's identity / Euclid's algorithm to equations in which b is divisible by a; for example, any equation in which a = 1.

For example, x + 2y = 5.

In applying Bézout's identity to this, I'd have to find h and k such that

1h + 2k = 1 (since 1 = GCF(1,2) ), right?

However, I don't think a single couplet (h, k) solves this kind of equation.

(Indeed, the problem arises when applying Euclid's algorithm to 1 and 2, because 2 is divisble by 1, so the algorithm ends right away.)

For a simple equation like x + 2y = 5, I think one might find all appropriate (h,k) couplets through trial and error relatively easily, but is there a more systematic way to solve these kinds of Diophantine equations that I could also apply to more complex problems?

In particular, how would I solve something like x + 3y + 5z = 3?

Thanks!

Gae Agn
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  • I'd try searching for earlier questions on this site, e.g., here. I don't think what you're asking will be easy to enumerate! (Certainly you can try different value of e.g. $z$ and use your earlier techniques to solve for $x$ and $y$...) – Benjamin Dickman Aug 24 '24 at 18:06
  • See the linked dupe for general algorithms. – Bill Dubuque Aug 24 '24 at 18:21

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