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I am doing maths for fun and stumbled upon this amazing worksheet. The second last question is a Diophantine equation with three variables and the solution ends with the general solution to the equation. I understand everything except how they go from

$$w-6z=2\\w=8+6k\\z=1+k\\x+3y=8+6k$$

to the "general solution of x and y" from the above equation to

$$x=8+3t\\y=2k-t$$

I understand that they get $w=8+6k$ and $z=1+k$ from the solution $(8, 1)$ but how they get to the general solution of x and y is unclear to me.

Alessio K
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linker
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3 Answers3

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$x+3y=8+6k$ let $y=t$, then $x=8+6k-3t$, where $t \in \Re$ So the solutions are is $x=8+6k-3t, y=t$.

Z Ahmed
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If you put $y = 2k - t$ in $x+3y=8+6k$, you get $x=8+3t$.

So, $(x,y) = (8+3t, 2k-t)$.

vvg
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You have $$2x+6y-12z=4$$or$$2(x+3y)-12z=4.$$

So setting $w=x+3y$ we have $$2w-12z=4$$ $$w-6z=2$$ which is the diophantine equation in two variables. The above has general solution $$w=8+6k$$$$z=1+k$$ for $k\in\mathbb Z$. Since substituting gives $w-6z=(8+6k)-6(1+k)=8-6+6k-6k=2$ as required. But we know that $w=x+3y$ so it follows that $$x+3y=8+6k.$$

So choosing $$x=8+3t$$ and $$y=2k-t$$ we see that it indeed satisfies the above equation.

This is called parametrisation. So for instance we can write the line $y=x+2$ to be $$(x,y)=(t,t+2)$$ for all $t\in\mathbb Z$ by setting $x=t$. From above we had $x+3y=8+6k$, now set $x=8+3t$ then we obtain $$x+3y=8+3t+3y=8+6k$$ $$3t+3y=6k$$ so $y=2k-t$.

Alessio K
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