# How to convert a diophantine equation into parametric form?

Let be $\space 7x+9y=5 \space$ a linear diophantine equation, in two variables. What are the integer solutions for $x$ and $y$?

I know that $7x+9y=5$ is a cartesian equation for a line in the plane. Then I thought, if one could define $x$ and $y$ in terms of the same parameter, it would be possible to know all the integer solutions. But I don't have clue on how can convert an diophantine equantion in the form $ax+by=c$ to its parametric form.

Can you give me some hints?Thanks.

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$\frac 97=1+\frac27=1+\frac1{\frac72}=1+\frac1{3+\frac12}$

So, the last but one convergent is $1+\frac13=\frac43$

Using Convergent property of continued fraction, $7\cdot4-9\cdot3=1$

$7x+9y=5(7\cdot4-9\cdot3)\implies 7(x-20)=-9(y+15)\implies x-20=\frac{-9(y+15)}{7}$ which is an integer.

So, $7\mid(y+15)$ as $(7,9)=1$ $\implies \frac{x-20}{-9}=\frac{y+15}7=z$ for some integer $z$

So, $y=7z-15=7(z-3)+6=7w+6$ where $w=z-3$ is any integer.

So, $x=-9z+20=-9(w-3)+20-27=-(9w+7)$

Alternatively, by observation $7x+9y=5=14-9$

or $7(x-2)=-9(y+1)$

or, $\frac{x-2}{-9}=\frac{y+1}7$

$\frac{-9(y+1)}7=x-2$ which is an integer, so is $\frac{y+1}7$ as $(7,9)=1$

So, $\frac{x-2}{-9}=\frac{y+1}7=u$ where $u$ is any integer.

So, $x=-9u+2,y=7u-1$

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Thanks for your awswer and sorry for the delay in my replay. I've been studying this theme.

Well, for the example we have $\space 7x+9y=5 \space$. Writing in the reduce form, one obtain $\space y=-\frac{7}{9}x+\frac{5}{9} \space$

Now the goal is to find a integer for $x$ that made $y$ integer too. In this example $5-7x$ must be a multiple of $9$. So it's easy to see that $x=2$ will fit.

Now, $y=-1$ and $x=2$ it's a particular solution.

I imagined the graph of this line, and as the line rise down, the $x$ runs to the right. By the reduce form of the line equation I know that the slope is $\space -\frac{7}{9} \space \left ( \frac{rise}{run} \right )$ . So the parametric equation would be:

$x=2+9t$

$y=-1-7t \space$, for $t \in \mathbb{Z}$.

If the slope was positive, these equations will be:

$x=2-9t$

$y=-1+7t \space$, for $t \in \mathbb{Z}$.

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Dear Joao, In the second formula, you should have $2 - 9t$. (You have substituted $-t$ for $t$ in the first formula.) Regards, –  Matt E Dec 17 '12 at 16:43
You are right. Making the same thought "as the line rise down, the x runs to the right".Now the slope is positive, I must say as the line rise up, the x runs to the left (decrease). –  João Dec 17 '12 at 17:01