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yesterday vote accept Tong Lingling
yesterday history edited Tong Lingling CC BY-SA 4.0
improve formatting
yesterday comment added Tong Lingling @Luna'sChalkboard I have verified the case for $k\leqslant 10^5$ and found no counterexamples. I will update in this post after extending the verification to a larger range.
yesterday history became hot network question
yesterday comment added David E Speyer I answered a similar question at mathoverflow.net/questions/442497 .
yesterday answer added Henri Cohen timeline score: 8
2 days ago comment added Tong Lingling @DanielWeber Sorry for my typo. '$r$ divides $m$' should be replaced with '$r$ divides $j$'.
2 days ago history edited Tong Lingling CC BY-SA 4.0
correct typo
2 days ago comment added Tong Lingling @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.
2 days ago history edited Tong Lingling CC BY-SA 4.0
supplement content
2 days ago comment added EffingLoveMath 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.
2 days ago comment added Daniel Weber 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 $?
2 days ago history edited Tong Lingling CC BY-SA 4.0
correct
2 days ago history edited Tong Lingling CC BY-SA 4.0
delete tag 'prime'
2 days ago history asked Tong Lingling CC BY-SA 4.0