Can one find all positive integer triplets $(x,y,z)$ satisfying this parametric equation :

$$ax^2 + (1-a)x + by^2 + (1-b)y = cz^2 + (1-c)z$$

Here, $a, b, c$ are positive integers greater than $1$.

$a≡1(mod$ $x)$

$b≡1(mod$ $y)$

$c≡1(mod$ $z)$

One solution is to write this equation in the form below? $$X(X+A)+Y(Y+A)=Z(Z+A)$$ See Invid's answser : https://mathoverflow.net/questions/142938/is-there-an-algorithm-to-solve-quadratic-diophantine-equations


Please see my answer at MO. To quote myself:

If $c=0$, one has: $$z=ax^2+by^2-ax-by+x+y$$ Which is obviously an integer.

If $c\neq0$, then, we can rewrite the equation as $$cz^2+(1-c)z-\left(ax^2+(1-a)x+by^2+(1-b)y\right)=0\\ \implies z=\dfrac{c-1\pm\sqrt{c (4 x (a (x-1)+1)+4 y (b (y-1)+1)+c-2)+1}}{2c}$$ By using the quadratic equation. We may set $x,y$ as integers. Then, $\left(x,y,\dfrac{c-1\pm\sqrt{c (4 x (a (x-1)+1)+4 y (b (y-1)+1)+c-2)+1}}{2c}\right)$ is the general form of the integer solution of the above equation ($c\neq0$) iff ${2c}\mid \left({c-1\pm\sqrt{c (4 x (a (x-1)+1)+4 y (b (y-1)+1)+c-2)+1}}\right)$. Any suggestions on simplifying this will be welcome.

  • $\begingroup$ Shouldn't this be closed as duplicate? I'm assuming that the OP are the same, since both are kiva2000 $\endgroup$ – Calvin Lin May 17 '14 at 1:25
  • 1
    $\begingroup$ @CalvinLin It should be closed as duplicate, however you can't close $M.SE$ questions as duplicates of $MO$ questions. $\endgroup$ – user122283 May 17 '14 at 1:29
  • $\begingroup$ This is such a joke answer perhaps? $\endgroup$ – individ May 17 '14 at 8:50
  • $\begingroup$ Thanks for the clarification, I wasn't aware that we couldn't close as duplicates of MO. $\endgroup$ – Calvin Lin May 17 '14 at 17:38

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