Timeline for sums of rational squares

7 events

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Dec 8, 2018 at 6:49	comment	added	supriya saha		Thank you very much for pointing out gcd(w,p)=1. From gcd(c,d)=1, gcd(c^{2},d^{2)=1. Thus, we can say k does not divide $c^{2}$ as $d^{2}=kb^{2}$. Then k=1 as it must give integer in the right-hand side. But your method is much better.
Dec 8, 2018 at 6:20	history	edited	supriya saha	CC BY-SA 4.0	added 16 characters in body
Dec 7, 2018 at 19:51	comment	added	user44191		Welcome to MathOverflow. I've edited your post to make clear the question you are answering. You may want to make clear that $gcd(w, p) = 1$ (as otherwise, $3^2 + 0 = 3*3$ is a counterexample to your statement). I'm not sure I understand your proof that $k = 1$; isn't it enough to show that $b^2 \| d^2$ and $d^2 \| b^2$? Further, you need that $gcd(a^2, b^2) = 1$ to show $b^2 \| d^2$.
Dec 7, 2018 at 19:47	history	edited	user44191	CC BY-SA 4.0	Clarifying the question being answered
Dec 7, 2018 at 19:45	review	Late answers
Dec 7, 2018 at 19:51
Dec 7, 2018 at 19:30	review	First posts
Dec 7, 2018 at 20:33
Dec 7, 2018 at 19:29	history	answered	supriya saha	CC BY-SA 4.0