0
$\begingroup$

How to determine the coefficient signs for the solution of a linear diophantine equation?

Take $24x + 69y = 33$ for example.

I know the solution is $x = 33 − 23k , y = −11 + 8k$, and I understand fully how to arrive at the values, but how do I determine the sign on the coefficients?

In other words, why is $23k$ negative, and why is $8k$ positive?

Through all the examples I can find, there seems to be no pattern.

$\endgroup$
0
$\begingroup$

You want to parameterize the solutions to the equation$$24x + 69y = 33$$

You know that $(x_0, y_0) = (33, -11)$ is a solution.

\begin{align} 24(\phantom{3}x) + 69(\phantom{-1}y) &= 33 \\ 24(33) + 69(-11) &= 33 &\text{(subtract)} \\ \hline 24(x-33) + 69(y+11) &= 0 \end{align}

Since $24$ divides $24(x-33)$, then it must also divide $69(y+11)$.

Since $24$ and $69$ are relatively prime to each other, then we must have $24$ divides $y+11$. So, for some integer $t, \ y+11 = 24t$. Hence $y=24t-11$. We can now solve for $x$.

\begin{align} 24(x-33) + 69(y+11) &= 0 \\ 24(x-33) + 69(24t-11+11) &= 0 \\ 24(x-33) + 69(24t) &= 0 \\ 24(x-33) &= -69(24t) &\text{Note the change in sign.} \\ x-33 &= -69t \\ x &= -69t + 33 \end{align}

And you get \begin{align} x &= -69t + 33 \\ y &= 24t - 11 \end{align}

If you let $t=-u$, you get \begin{align} x &= 69u + 33 \\ y &= -24u - 11 \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.