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Let $a,b,c$ be positive integers. Verify that Diophantine equation $ax + by = c$ has integer solution $x_0, y_0$ if and only if $GCD(a,b)|c$.

Attempt

Diophantine $ax + by = c$ has integer solution $x_0, y_0$ implies $GCD(a,b)|c$. Factoring out $ GCD(a,b)$ from each side gives $$\frac{1}{GCD(a,b)}(ax + by) = \frac{1}{GCD(a,b)}c$$ which must still have an integer solution as $GCD(a,b)$ obviously divides both $a$ and $b$. If $GCD(a,b)$ does not divide $c$ there is no integer solution.

That $GCD(a,b)|c$ implies that $ax + by = c$ has integer solution $x_0, y_0$. Let $c = GCD(a,b)d$, where $d$ is integer.

$ax + by = GCD(a,b)d$; divide both sides by $GCD(a,b)$: Obtain $mx + ny = d$ with $GCD(m,n) = 1$. If $d$ is an integer and $GCD(m,n) = 1$ then $x,y$ must be integers as $mx + ny$ cannot be further divided evenly.

or

$ax + by = c$ has no solution implies $GCD(a,b)$ does not divide c.
$c = GCD(a,b)(\frac{ax}{GCD(a,b)} + \frac{by}{GCD(a,b)}) + r$

with $GCD(a,b) > 0, 0<r<GCD(a,b)$, r integer

$\frac{c}{GCD(a,b)} = \frac{ax + by}{GCD(a,b)} + \frac{r}{GCD(a,b)}$

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  • $\begingroup$ It seems like you are attempting to prove the direction $(\Rightarrow)$. You should say more to make the reasoning clear. For the other direction you can use Bezout's identity for the gcd $\endgroup$ – Bill Dubuque Jul 28 '14 at 19:21
  • $\begingroup$ I don't think I need to use Bezout's as that is not mentioned in the chapter. Maybe the underlying principles. $\endgroup$ – miniparser Jul 29 '14 at 4:43
  • $\begingroup$ Bezout's identity for the gcd is the special case $\,c = \gcd(a,b),\,$ of the direction $(\Leftarrow),\,$ so you will need Bezout (or some equivalent). $\endgroup$ – Bill Dubuque Jul 29 '14 at 4:51
  • $\begingroup$ I reformatted the question. Please see math notation guide. $\endgroup$ – user147263 Jul 30 '14 at 21:36
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If $(a,b)\mid c$, $\exists t\in \mathbb{Z}$ such that $t(a,b)=c$. As we know that there exists $x,y \in \mathbb{Z}$ such that $ax+by=(a,b)$, then choose the integers $x_0=tx$ and $y_0=ty$.

To prove the converse, if $x_0$ and $y_0$ be integers, then $(a,b)\mid x_0a$ and $(a,b)\mid y_0y$ implies $(a,b)\mid (x_0a+y_0b)$.

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  • $\begingroup$ Wow, this is much more concise that what I stated. $\endgroup$ – KernelPanik Aug 11 '14 at 12:20
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Let a,b,c be positive integers. Verify that Diophantine equation ax+by=c has integer solution x0,y0 if and only if GCD(a,b)|c.

=>: (how do you write the arrows?):

This was pretty much correct in my first attempt. More clearly:

If $ax + by = c$ has integer solution $x_0,y_0$(they exist - how do i draw that?) then $ax_0 + by_0 = c$. Divide both sides by $gcd(a,b)$. (See first attempt.) Then $gcd(a,b)|a, gcd(a,b)|b$ => $gcd(a,b)|(ax_0 + by_0)$ => $gcd(a,b)|c$.

<=:

This was pretty much wrong in my first attempt.

Let $gcd(a,b)|c$

Then exist e such that $c = gcd(a,b)*e$ and exist $x_1,y_1$ with $gcd(a,b) = ax_1 + by_1$

Multiply both sides by $e$: $gcd(a,b)*e = ax_1e + by_1e$

$a(x_1e) + b(y_1e) = c$, $(x_1e,y_1e)$ solution.


Attempt to prove both ways at once:

$ax + by = c$ has integer solution $ax_0 + by_0 = c$

Let $d = gcd(a,b)$; then $a = md,b = nd$:

$c = (md)x_0 + (nd)y_0 = d(mx_0 + ny_0)$

So $d$ divides $c$.

If either $ax_0 + by_0 <>($does not equal$) c$ or $gcd(a,b) $ does not divide $ c$

$c = (md)x_0 + (nd)y_0 = d(mx_0 + ny_0)$ would not be valid. kind of a stretch there.

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Let $d=GCD(a,b)$ where $ {a,b,c,d,k_1,k_2,x,y} \in \mathbb Z$

(a) Positive: $d|c$

$$ d|a \implies d|ax \\ d|b \implies d|by $$

Since by the Division Theorem

$$ d|a \implies a/d=k_1\\ x(a/d)=k_1x \implies d|ax \\ d|b \implies b/d=k_2\\ y(b/d)=k_2y \implies d|by $$

Since $d|ax$ and $d|by$, and there exists a solution to $ax + by = c$, then $d$ divides both sides of the equation $ax + by = c$ and therefore $d|c$.

(b) Contrapositive: $d\not|c$

Given $d|a$ $$ d\not|ax \implies x\notin \mathbb Z $$ Likewise, given $d|b$ : $$ d\not|by \implies y\notin \mathbb Z $$

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