Here's a question about coprimes that I stumbled upon while doing some research. Providing insight into this question would prove quite helpful to me.
Choose a pair of coprimes $x, y \in \mathbb Z$. What constraints have to be placed on $x$ and $y$ for the following pair of equations to have integral solutions?
$$ax + by = 1$$ $$a + b = 0$$
The first equation can be solved from the assumption that $(x, y) = 1$. It is not obvious to me when the second constraint can be satisfied, if ever.
I've restated the problem a bit using insights from Bézout's identity and the equations describing the general structure of Bézout coefficients.
First, let $k$ be the difference bewteen $x$ and $y$ so that $k = x - y$. $y$ can then be expressed in terms of $x$ with the identity $y = x - k$. The equations can then be restated:
$$ax + b(x-k) = 1$$ $$a + b = 0$$
The coefficients $a$ and $b$ which satisfy the first equation are not unique. If the Bézout coefficients (a, b) provide a solution to the equation
$$ax + b(x-k) = 1$$
then for any $m \in \mathbb Z$, the following coefficients also provide a solution:
$$(a + m(x-k),\; b - mx)$$
The constraint that $a + b = 0$ can be generalized to all coefficients:
$$(a + m(x-k)) + (b - mx) = 0 \implies$$ $$a + mx - mk + b - mx = 0 \implies$$ $$a + b - mk = 0 \implies$$ $$a + b \;|\; k$$
Substituting $x - y$ for $k$ we then require
$$a + b \;|\; x - y$$
or
$$ a + b \equiv 0 \;\;\;mod\;\; x - y$$
Now the problem can be restated in terms of division. For coprimes $x$ and $y$, when can Bézout coefficients be found satisfying $a + b \;|\; x - y$, where
$$ax + by = 1$$
Said another way, what constraints must be places on coprimes $x$ and $y$ so that they posess a pair of Bézout coefficients whose sum divides their difference?
Any insights would be much appreciated.