MHB How do I solve the Diophantine equation $4x+51y=9$?

[Ackbach]

Here is this week's POTW:

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Solve the Diophantine equation $4x+51y=9$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to kaliprasad for two correct solutions. I post one of them here for you:

Spoiler

Because $\gcd(51,4)=1,$ a solution exists for the same.

Find $1$ as a combination of $51$ and $4$ using the Extended Euclidean Algorithm:

$51=12∗4+3$ or $3=51−12∗4$

$4=3∗1+1$ or $1=4−3=4−(51−12∗4)=13∗4−51∗1$

So $9= 4 * 2 + 1 = 4 * 2 + (13 * 4 - 1 * 51) = 15 * 4 - 1 *51$.

So $x = 15, y = -1$ is a solution of the same, as $51 * 4 - 4 * 51 = 0$; so adding $51t$ to $x$ and subtracting $4t$ from $y$ will not change $4x+51y$. So $x=15+51t$, $y=−1−4t$ is the solution set, where $t$ is any integer.