Prove the required form

Question

Let $a$ and $b$ be coprime integers and $(x_0, y_0)$ be a set of integer solutions of the Diophantine equation $ax+by=1$. Prove that any set of integer solution is of the form $x = x_0 + bt$, $y = y_0 - at$, where $t\in\Bbb{Z}$.

Answer 1

Assume $a$ and $b$ are both nonzero.

Consider $G=\{(x,y)\in\mathbb{Z}^2:ax+by=0\}$. This is clearly a proper and nontrivial subgroup of $\mathbb{Z}^2$, so it is generated by a single element, namely $(b,-a)$.

However, the general theorem can be avoided in this simple case. Indeed, if $ax+by=0$, then $$ ax=-by $$ so $a\mid by$ and, since $\gcd(a,b)=1$, we get $a\mid y$. Therefore $y=as$. Similarly, $x=bt$. Now $abt+abs=0$ implies $s=-t$, so $$ (x,y)=(bt,-at)=t(b,-a) $$ Conversely, any element of the form $t(b,-a)=(tb,-ta)$ is a solution.

Suppose $(x_0,y_0)$ and $(x_1,y_1)$ are solutions of $ax+by=1$. Then $(x_1-x_0,y_1-y_0)$ is a solution of $ax+by=0$, so $$ (x_1,y_1)=(x_0,y_0)+t(b,-a) $$ and, conversely, any pair of this kind is a solution of $ax+by=1$. Therefore all solutions of $ax+by=1$ are of the form $$ x=x_0+tb,\qquad y=y_0-at $$

Suppose $b=0$, so $a=\pm1$. For $a=1$, the solutions of $ax+0y=1$ are of the form $x=1+0t$, $y=0-1t$. Similarly if $a=-1$.

How to find a particular solution $(x_0,y_0)$ is another matter.

Answer 2

We assume there is a solution $x,$ $y$ with $ax+by=1.$ We define $t=\frac{x-x_0}{b}$ (not necessarily integer), which is equivalent with $t=\frac{y_0-y}{a}.$ We have $bt=x-x_0$ and $at=y_0-y.$

Then $t = (ax_0+by_0)\,t = at\,x_0+bt\,y_0 = (y_0-y)x_0 + (x-x_0)y_0 = xy_0-yx_0 \in\mathbb{Z},$ and we obtain $x=x_0+bt$ and $y=y_0-at$ with $t\in\mathbb{Z}.$

Answer 3

Subtract $ax_0+by_0=1$ from $ax+by=1$ ?

Answer 4

\begin{align} ax + by &=1 \\ ax_0 + by_0 &= 1 \\ \hline a(x-x_0) + b(y-y_0) &= 0 \\ b(y-y_0) &= -a(x-x_0) \tag{1.} \end{align}

Since $a$ and $b$ are coprime, then there exists integers $A$ and $B$ such that $aA +bB = 1$. Hence \begin{align} aA +bB &= 1 \\ a(y-y_0)A + b(y-y_0)B &= (y-y_0) \\ a(y-y_0)A - a(x-x_0)B &= (y-y_0) &\text{(Using (1.))} \tag{2.}\\ \end{align}

Clearly $a$ divides the left side of equation $(2.)$. Hence $a$ divides $y-y_0$, say

$$y-y_0 = at \tag{3.}$$

for some integer $t$. Substituting this back into equation $(1.)$, we get $bat = -a(x-x_0)$. Dividing both sides by $a$, we get

$$x-x_0 = -bt \tag{4.}$$

Solving equations $(3.)$ and $(4.)$ for $x$ and $y$, we find that, for some integer t, $$ x=x_0-bt \quad \text{and} \quad y=y_0+bt$$

It works the other way too. $ax + by = a(x_0 - bt) + b(y_0 + bt) = ax_0 + by_0 = 1$

So $(x,y)$ is a solution to $ax+by=1$ if and only if for some integer $t$, $ x=x_0-bt \quad \text{and} \quad y=y_0+bt$.