acsuvaic9

2022-07-17

Solve diophantine equation ${x}^{2}-2{y}^{2}=x-2y$

Selden1f

Expert

The equation is
${x}^{2}-x-2\left({y}^{2}-y\right)=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{\left(x-\frac{1}{2}\right)}^{2}-2{\left(y-\frac{1}{2}\right)}^{2}-\frac{1}{4}+\frac{1}{2}=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{\left(x-\frac{1}{2}\right)}^{2}-2{\left(y-\frac{1}{2}\right)}^{2}=-\frac{1}{4}$
Now substitute
${x}^{\prime }:=x-\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime }:=y-\frac{1}{2}$
So that the equation becomes
${x}^{\prime 2}-2{y}^{\prime 2}=-\frac{1}{4}$
and it now has the form you're used to.

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