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Sep 19, 2020 at 21:01 comment added Timothy Chow I should remark that $x$ could be slightly larger than $\sqrt p$ if $y$ is negative. But if $p = x^2 - xy + y^2 = (x-y)^2 + xy$ for some $x>0$ and $y>0$ then $|x-y| < \sqrt p$. So neither $x$ nor $y$ can be much larger than $\sqrt p$, because then $xy$ would be larger than $p$.
Sep 16, 2020 at 18:31 vote accept Gautam
Sep 16, 2020 at 0:27 history edited Timothy Chow CC BY-SA 4.0
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S Sep 15, 2020 at 22:58 history answered Timothy Chow CC BY-SA 4.0
S Sep 15, 2020 at 22:58 history made wiki Post Made Community Wiki by Timothy Chow