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Given $a,b\in\mathbb{Z}-\{0\}$ and $N\in\mathbb{Z}$, it is easy to show that if $x_0,y_0\in\mathbb{Z}$ are a particular solution to $ax+by=N$, then $x=x_0+\frac{b}{d}t$ and $y=y_0-\frac{a}{d}t$, where $d=gcd(a,b)$ and $t\in\mathbb{Z}$, are also solution to $ax+by=N$.

But may I ask how to proof that they are actually the general solution to $ax+by=N$ if we restrict the solutions inside $\mathbb{Z}$? (i.e. all integer solutions have been counted)

Thanks!

J-A-S
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  • Suppose that $x_1,y_1$ is another solution, then $ax_0-ax_1+by_0-by_1=0$ and therefore $a\mid by_0-by_1$ as well as $b\mid ax_0-ax_1$. This can be morphed further to show that the solution equations for $x,y$ must hold. – abiessu Dec 26 '20 at 05:47
  • @abiessu sorry I don't really see how that proves the generality. Maybe I'm a bit new to this topic, could you please be more specific? thanks – J-A-S Dec 26 '20 at 07:26
  • If the distance vector between any two solutions can be shown to always be divisible by the vector $(\frac bd,\frac ad)$ where $d=$gcd$(a,b)$, then there is always a $t$ which produces such solutions from the given solution equations for $x,y$. – abiessu Dec 26 '20 at 07:42

1 Answers1

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Let A be the set of all ordered pairs integer solutions. Let B be the set of all ordered pairs of integer solutions only of the form you gave. We know $B \subseteq A$

First find all the rational solutions for the equation, then restrict them.

Let

$x=x_0+bu$

for $u \in\mathbb{Q}$

This is solvable for u for any rational x.

And then using

$ax+by=N$

$a(x_0+bu)+by=N$

$y=\frac{N-a(x_0+bu)}{b}$

$y=\frac{by_0-abu}{b} = y_0-au$, which is also rational.

So every element of A can be written as $(x_0+bu,y_0-au)$ for some rational u.

So let $(x_0+bu,y_0-au) \in A$

We require

$bu \in \mathbb{Z}$

$au \in \mathbb{Z}$

write $u=\frac{m}{n}$. Assume this is in lowest terms

So

$\frac{bm}{n} \in \mathbb{Z}$

$\frac{am}{n} \in \mathbb{Z}$

So $n|b$ and $n|a$

That means $n|d$ where $d=gcd(a,b)$

We can write $rn=d$ for some integer r

So $n = \frac{d}{r}$

$\frac{bm}{n} = \frac{b}{d}(rm)$

$\frac{am}{n} = \frac{a}{d}(rm)$

So letting $t=rm$, we know that $(x_0+\frac{b}{d}t,y_0-\frac{a}{d}t) \in B$

So $A \subseteq B$ giving us $A=B$.

Ameet Sharma
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