# Linear Diophantine Equations

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• Jun 25th 2010, 12:26 PM
dwsmith
Linear Diophantine Equations
$\displaystyle 12x+13y=14$

$\displaystyle gcd(12,13)=1\rightarrow \ 1|14$

Can the equations x and y be found using the vector $\displaystyle <12,16> \ \mbox{and} \ (-1,2) \mbox{?}$

$\displaystyle x=-1+12t$
$\displaystyle y=2+13t$

Does this work?
• Jun 25th 2010, 02:05 PM
jamix
I'm not sure I understand your two vectors $\displaystyle <12,16>$ and $\displaystyle <-1,2>$.

It is true that $\displaystyle (x_0,y_0) = (-1,2)$ is a solution to the equation $\displaystyle 12x + 13y = 14$ and that the general solution is of the form $\displaystyle (x_0 + a \cdot t, y_0 + b \cdot t)$ where $\displaystyle (a,b)$ are integer constants, however in this case the constants are not $\displaystyle (12,13)$ but rather $\displaystyle (13,12)$.

Can you see why?

Hence the general solution is $\displaystyle x = -1 + 13 \cdot t$ and $\displaystyle y = 2 + 12 \cdot t$
• Jun 25th 2010, 02:20 PM
dwsmith
There aren't two vectors. That is why I said vector not vectors. The vector is denoted $\displaystyle <,>$ and $\displaystyle (,)$ denotes a point.

You know how if you have a vector and a point you can parametrize it? Can that be done for these types of equations?
• Jun 25th 2010, 03:16 PM
jamix
It depends on whether or not the point is a solution the equation.

Regardless of the point used as your initial solution $\displaystyle (x_0,y_0)$, there is only ONE vector $\displaystyle v = <v_1,v_2>$ that can be used to describe the entire set of solutions as $\displaystyle (x_0 + v_1, y_0 - v_2)$

Is this what you wanted?
• Jun 25th 2010, 06:49 PM
chiph588@