1

I hope I have provided all the information necessary to make my question clear and understandable.

The following theorem is well known in elementary number theory

Theorem. The linear Diophantine equation $ax+by = c$ has a solution if and only if $\, d \vert c \,$ where $d = \text{gcd}(a,b)$. If $x_0$,$y_0$ is any particular solution of this equation, then all other solutions are given by

$$x = x_0 + \bigg(\frac{b}{d}\bigg)t$$ $$y = y_0 - \bigg(\frac{a}{d}\bigg)t$$

where $t$ is an arbitrary integer (bold emphasis added).

Question. How is the transition made from $t$ being an existential integer in the proof to it being an arbitrary integer at the end of the proof?

Proof. We begin with two solutions $(x_0,y_0)$ and $(x\prime, y\prime)$. Then starting with the equation

$$ax_0 + by_0 = c = ax\prime + b\prime$$

we have

$$r(x\prime - x_0) = s(y\prime - y_0)$$

where $r$ and $s$ are relatively prime integers from $a = dr$ and $b = ds$. Consequently, Euclid's Lemma allows us to deduce that $r \vert (y_0 - y\prime)$; or $y_0 - y\prime = rt$ for some integer $t$.

So now we have some integer $t$ such that

$$y_0 - y\prime = rt$$

Substituting, we obtain

$$x\prime - x_0 = st$$

We then have the formulas

$$x\prime = x_0 + st = x_0 + \bigg(\frac{b}{d}\bigg)t$$ $$y\prime = y_0 + rt = y_0 - \bigg(\frac{a}{d}\bigg)$$

But then we show that the integer $t$ has no restrictions as to its values, because

$$\begin{align} ax\prime+by\prime &= a\bigg[x_0+\bigg(\frac{b}{d}\bigg)t\bigg] + b \bigg[y_0 - \bigg(\frac{a}{d}\bigg)t\bigg] \\ &= a(x_0+by_0) + \bigg(\frac{ab}{d} - \frac{ab}{d}\bigg)t \\ &= c + 0 \cdot t \\ &= c \end{align} $$

There are an infinite number of solutions of the given equation, one for each value of $t$

$\blacksquare$

I'm confused as to how $t$ can become arbitrary when it is existential. How do we extend the quantifier from existence to universality?

J. Dunivin
  • 3,103
  • The solutions of this diophantine equation have a special structure. If we have any special solution $s$ of $ax+by=c$ and an arbitary solution of the homogenous equation $ax+by=0$, lets call it $r$ , then $s+r$ is a solution of $ax+by=c$ as well. And the homogenous solution is $(-bt/at)$ , $t\in \mathbb Z$ , if $a$ and $b$ are coprime. – Peter May 07 '17 at 16:53
  • So, if we have a solution, there are infinite many of them and we can parametrize it such that every solution corresponds with an integer number $t$ – Peter May 07 '17 at 16:57
  • If I understand what you're saying, because $t$ doesn't affect the solution, we can introduce an arbitrary integer $r$, and replace $t$ with it, which would be parameterized. – J. Dunivin May 07 '17 at 22:58
  • 1
    Until you demonstrated that it was arbitrary, it was only existential. You can consider it to be a happy coincidence or a sign of some deeper, as yet undiscovered, property of numbers. If you have ever used the slope of a line to find more points on that line, this result shouldn't come as a suprise to you. – Steven Alexis Gregory Aug 17 '17 at 06:38
  • The first half of the proof shows that if $(x_0,y_0)$ is any particular solution, then any other possible solution must have form $(x,y) = (x_0,y_0) + t(x_h,y_h),,$ for $,t\in\Bbb Z.,$ The second half verifies that all of those possible solutions (all $,t\in \Bbb Z)$ are actually solutions. (Here $,(x_h,y_h) = (b/d,-a/d),\ d = \gcd(a,b)$ is the solution of the associated homogenous equation) – Bill Dubuque Aug 10 '23 at 17:09

0 Answers0