I think I might be misunderstanding how to solve diophantine equations. I seem to be making a mistake somewhere in my process, and I am having difficulty figuring out what it could be. Here is what I did:
$18x + 28y = 10$
gcd(18,28) = 2 and 2|10 so there are infinitely many solutions.
Reduce equation to $9x + 14y = 5$
let $x = -1$, $y = 1$
$9(-1) + 14(1) = 5$
$5 = 5$
And so we have one possible $x_0,y_0$
Plugging this into the following:
$x = x_0 + \frac{b}{d}n$ and $y = y_0 – \frac{a}{d}n$
we have all integral solutions are, $\forall n \in \mathbb{Z}$
$x = -1 + 14n$ and $y = 1 – 9n$
But the answer is supposed to be $x = -1 – 14n$, $y = 1 + 9n$. Am I possibly missing a step?
I was using the following theorem to solve this:
Let $ax + by = c$ be a linear Diophantine equation in the
two variables $x$ and $y$ and let d = gcd(a,b). If $d \nmid c$, then the equation has no solutions; if $d|c$, then the equation has infinitely many solutions. Furthermore, if $x_0$, $y_0$ is a particular solution of the equation, then all solutions are given by
$x$, $y$ where $x = x_0 + (\frac{b}{d})n$ and $y = y_0 – (\frac{a}{d})n$ where $n \in \mathbb{Z}$.
Best Answer
You have obtained the solution set $$ \begin{cases} x = -1 + 14n \\ y = 1 - 9n. \end{cases} $$ Writing the solutions as ordered pairs $(x, y)$, the solutions are $$ \ldots, \underset{n = -2}{\underbrace{(-29, 19)}}, \ \underset{n = -1}{\underbrace{(-15, 10)}}, \ \underset{n = 0}{\underbrace{(-1, 1)}}, \ \underset{n = 1}{\underbrace{(13, -8)}}, \ \underset{n = 2}{\underbrace{(27, -17)}}, \ldots. $$
Now let's consider the solution set $$ \begin{cases} x = -1 - 14n \\ y = 1 + 9n. \end{cases} $$ According to this, the solutions would then be $$ \ldots, \underset{n = -2}{\underbrace{(27, -17)}}, \ \underset{n = -1}{\underbrace{(13, -8)}}, \ \underset{n = 0}{\underbrace{(-1, 1)}}, \ \underset{n = 1}{\underbrace{(-15, 10)}}, \ \underset{n = 2}{\underbrace{(-29, 19)}}, \ldots. $$ But this is the same as the list you obtained, just written in the opposite order!