Skip to main content
MathOverflow

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

6
  • 2
    $\begingroup$ Could you explain how to get from $\frac{1}{2}\sum_{j=1}^{2k}\left(\left\lfloor\frac{j^2}{m}\right\rfloor+\left\lfloor\frac{j'^2}{m}\right\rfloor\right)$ to $\frac{2k(k-1)}{3}+\frac{1}{2}\left\lfloor\frac{2k}{r}\right\rfloor $? $\endgroup$ Commented 2 days ago
  • 1
    $\begingroup$ Why is your set of test cases so small? Usually, we test all $k = 1...N$ and numbers $m=4k+1$ that meet the conditions, where $N = 20...10,000$ the higher the more confident we can be to proceed with the proof assuming no counter-example was found. $\endgroup$ Commented 2 days ago
  • 2
    $\begingroup$ @DanielWeber I've edited and supplemented some steps. It makes use of the identity $\left\lfloor\frac{j^2}{m}\right\rfloor + \left\lfloor\frac{j'^2}{m}\right\rfloor = \frac{j^2}{m}+\frac{j'^2}{m}+\epsilon$, where $\epsilon=0$ if $r$ divide $m$, and $\epsilon=-1$ in other case. $\endgroup$ Commented 2 days ago
  • 1
    $\begingroup$ @DanielWeber Sorry for my typo. '$r$ divides $m$' should be replaced with '$r$ divides $j$'. $\endgroup$ Commented 2 days ago
  • 3
    $\begingroup$ I answered a similar question at mathoverflow.net/questions/442497 . $\endgroup$ Commented yesterday