# Absolute solution

$|x^2-xy-y^2|=1$ The solutions for the equation above are $$(x_1,y_1), (x_2,y_2), (x_3,y_3),\dots$$, where $$x$$ and $$y$$ are two natural numbers.

Which is bigger, $$A$$ or $$B$$? \begin{align} A & = x_1^2+x_2^2+x_3^2+x_4^2+\dots+x_{2017}^2 \\ \ \\ B & = y_{2018}\times y_{2019}\end{align}

Note: For any positive integer $$k$$, $$x_k\leq x_{k+1}$$.

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