Let's generalise it. Suppose we have 2 positive integers such that

We'll show that if then has solutions with nonnegative integers.

If are integers ( positive, negative or 0), then the equation: always has solutions. Indeed, by just considering there are solutions. ( multiply by the modular inverse of b)

Now suppose , we'll show that, with that restriction, the solution to is unique. ( It exists and is unique)

Suppose then however and so if is multiple of ( as it is), it must be thus and from there we immediately get so the solution is unique. For the existence just consider , since we can find satisfying that condition.

Now suppose again we only had the restriction , we know that in this case the solution for integers is unique. If then must be greater than Indeed, suppose it wasn't, then we haveABSURD!So we have found that if there's a solution in the nonnegative integers!

Now note that , so, by the unique of the representation if and x, y are integers, it follows that, if there can be no solution in the positive integers.

So thesmallestinteger such that for all ,has a solution for nonnegative integersis