# Linear Diophantine Equations

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

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

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

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

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

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

Can you see why?

Hence the general solution is $x = -1 + 13 \cdot t$ and $y = 2 + 12 \cdot t$
• Jun 25th 2010, 02:20 PM
dwsmith
There aren't two vectors. That is why I said vector not vectors. The vector is denoted $<,>$ and $(,)$ 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 $(x_0,y_0)$, there is only ONE vector $v = $ that can be used to describe the entire set of solutions as $(x_0 + v_1, y_0 - v_2)$

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