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How can I find all integer solutions to the equation

$$ x (1 + 3 x) - y(3 + 3 y) - z(4 + 3 z) = 2?$$

I'm familiar with Lagrange's approach to binary quadratic equations, but am unsure how to deal with the third variable that appears in the above equation. Any progress towards a solution to this particular problem or references to a more general theory would be very much appreciated.

user449881
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  • $x=X-1$ ; $x(3x+1)=(X-1)(3X-2)=X(3X-5)+2$ $$y(3y+3)+z(3z+4)=X(3X-5)$$ https://math.stackexchange.com/questions/794510/curves-triangular-numbers – individ May 26 '17 at 18:23
  • I would start by collecting all the factors $3$ together, then $x-z-2$ is a multiple of $3$. – Old Peter May 26 '17 at 19:01

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Multiply the equation by $12$ & arrange a little, we have \begin{eqnarray*} 36x^2+12x+1 =36y^2+36y+9 + 36z^2+48z+16 \\ (6x+1)^2=(6y+3)^2+(6z+4)^2 \end{eqnarray*} So it is just Pythagorus in disguise ... Eg $(3,4,5)$ gives $x=-1,y=0,z=0$ or ... more solutions can be generated using a deeper analysis modulo $6$ and using the usual parameterisation of pythagorean triples ... \begin{eqnarray*} 6x+1=-(a^2+b^2) \\ 6y+3=a^2-b^2 \\ 6z+4=2ab \end{eqnarray*} Now choose $a \equiv 2 \mod 6$ and $b \equiv 1 \mod 6 $ to generate more solutions ... Eg $a=8,b=1$ gives $(x,y,z)=(-11,10,2)$.

Donald Splutterwit
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