Generalizing a Pair of Diophantine Equations

Source abstract

For coprime integers $a$ and $b$, it is known that exactly one of the two Diophantine equations $$ax+by\ =\ \frac{(a-1)(b-1)}{2} \qquad\text{and}\qquad 1+ax+by\ =\ \frac{(a-1)(b-1)}{2}$$ admits a nonnegative integer solution, and that this solution is unique. We first generalize this result by replacing the right-hand side with an arbitrary integer $m$ and its complement $ab-a-b-m$. This framework enables us to study the existence and uniqueness of nonnegative integer solutions to $$ax+by\ =\ \frac{(a-1)(b-1)}{k} \qquad\text{and}\qquad 1+ax+by\ =\ \frac{(a-1)(b-1)}{k},$$ where $k$ is a fixed positive integer. We then obtain explicit results when $a$ and $b$ are consecutive Fibonacci numbers. Finally, we examine the original pair of equations in several particular settings, including when $b\equiv \pm1\mod a$, when $b$ is replaced by a higher power, and when the parameters are squared.