# How to solve quadratic Diophantine equation with 3 variables

Given the equation:

$3x^2 - x - 3y^2 + y = 3n^2 - n$

I'd imagine solving this involves techniques for solving Diophantines? Or am I wrong?

Could someone point me in the right direction?

• This becomes elaborate fairly quickly. Why do you want to know? Anyway, multiply through by $12$ and complete the squares. – Will Jagy Jan 3 '16 at 4:02
• @WillJagy I need to solve this equation as a piece in a computational problem. – Thevesh Theva Jan 3 '16 at 4:03
• math.stackexchange.com/questions/794510/… For this equation, one can dispense with the equation of Pell. But why are you asking questions? Still the answers don't interest you. – individ Jan 3 '16 at 4:24
• @individ thank you very much! What do you mean by the answers not interesting me? – Thevesh Theva Jan 3 '16 at 7:15

Too long for a comment:

• Renaming y and n with a and b, we have $f(x)=f(a)+f(b),$ where $f(t)=3t^2-t.$

• It is obvious that if $(a,b,x)$ represents a solution, then so does $(b,a,x).$

• Since $f(0)=0,\quad(0,x,x)$ and $(x,0,x)$ always constitute solutions.

• Therefore, it is enough to take into consideration the case $a\le b$ with $ab\neq0.$

• This being said, for $-100\le a\le b\le100,$ we have the following non-trivial solutions :

$\qquad\qquad\qquad\qquad$

• I offer this numerical data in the hope that it will aid future analytic answers.
• @individ: You can turn that into an answer, and check to see if by any chance there aren't other families of solutions, by comparing it to the above-given list. – Lucian Jan 3 '16 at 8:02
• The bug was. Already fixed. Well and what sense was to write all these numbers? That complex equation? $$3n^2-n+3y^2-y=3x^2-x$$ Had written such solution. $$n=18p^2+18ps+p+\frac{s(9s+1)}{2}$$ $$y=24p^2+24ps-2p+s(6s-1)$$ $$x=30p^2+30ps-p+\frac{s(15s-1)}{2}$$ – individ Jan 3 '16 at 8:03
• A more simple option. $$n=(6k+7)t$$ $$y=(18k^2+42k+24)t-k-1$$ $$x=(18k^2+42k+25)t-k-1$$ – individ Jan 3 '16 at 8:56