Preferably for when $x < y < z$, how can I find every integral solution for this equation: $$-9x^2-13x+18y^2+18y-9z^2-z=0$$ Some solutions are: $(x,y,z) = (0,25,36), (4,8,11), (9,14,18), (10,12,14), (11,21,28)$
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In General, the formula there. https://math.stackexchange.com/questions/794510/curves-triangular-numbers If the coefficients do not represent a trivial case, the solutions of this equation in integers is always there. – individ Aug 02 '17 at 04:19
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You might start by writing this as
$$ (18 z + 1)^2 + (18 x + 13)^2 - 2 (18 y + 9)^2 = 8 $$
and thus
$$ Z^2 + X^2 - 2 Y^2 = 8 $$ where $X = 18 x + 13 \equiv 13 \mod 18$, $Y = 18y+9 \equiv 9 \mod 18$, $Z = 18 z + 1 \equiv 1 \mod 18$. Note that this does work mod $18$.
For fixed $Z$, this is a Pell equation. If $\pmatrix{X\cr Y}$ is a solution, then so is $\pmatrix{3 & 4\cr 2 & 3\cr}^k \pmatrix{X \cr Y}$ for any $k$ (with the same $Z$). If $k$ is divisible by $12$, this will be the same mod $18$ as $\pmatrix{X\cr Y}$. Thus we will get infinite families of solutions.
Robert Israel
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