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Sep 10, 2016 at 21:23 comment added Wojowu @KConrad I see, and I agree. Having four squares in both numerator and denominator was just the first idea I had, but as you mention it does look overcomplicated if you realize you can deal with a single square in a denominator.
Sep 10, 2016 at 21:19 comment added KConrad You proved it yourself: start with the ideas for Lagrange and change the end of the argument a little bit. I disagree that your definition makes it "much clearer" that a reciprocal of a good number is good. For any nonzero sum of $k$ squares, $a = \sum_{i=1}^n a_i^2$, we have $1/a = a/a^2 = \sum_{i=1}^k (a_i/a)^2$ is a sum of $k$ squares. I stand by my belief that writing the denominator as a sum of four squares instead of as a single square makes the condition look overly complicated.
Sep 10, 2016 at 19:02 comment added Wojowu @KConrad It took me a little bit of thought, but I can agree with you. My definition of "good" makes it much clearer though why a reciprocal of good number is good. But I agree that argument with sums of four rational squares would be also be neat. Can you see a direct way (without Lagrange) to show that a good number (with my definition) is a sum of four rational squares?
Sep 10, 2016 at 18:40 comment added KConrad Your definition of "good" is overly complicated. It is the same thing as being a sum of four rational squares. I think the definition of "good" should be described that way and the proof you wrote continues to work with no major changes. This puts into better perspective how this approach differs from the one using Lagrange: it's enough to use sums of four rational squares rather than four integer squares (nontrivial work shows those are the same property, but you don't need such an equivalence for this proof).
Sep 10, 2016 at 13:14 vote accept CommunityBot
Sep 10, 2016 at 13:14 comment added user40023 Very nice idea.
Sep 10, 2016 at 10:38 history edited Wojowu CC BY-SA 3.0
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Sep 10, 2016 at 9:56 history answered Wojowu CC BY-SA 3.0