Is there finite number of $p$ such $\sum_{k=1}^{p-1}\left\{\frac{k^2}{p}\right\}=\frac{p-3}{2}？$

Question

Let $p$ be odd prime number, and $p\equiv 3\pmod 4$,Is there finite number of $p$ such $$\sum_{k=1}^{p-1}\left\{\dfrac{k^2}{p}\right\}=\dfrac{p-3}{2}？$$

where $\{x\}=x-[x]$

I known if $p\equiv 1\pmod 4$,then we have $\sum_{k=1}^{p-1}\left\{\dfrac{k^2}{p}\right\}=\dfrac{p-1}{2}$can see this 5-th Hong Kong Mathematical Olympiad 2002 problem and solution

score 1 · Answer 1 · answered Mar 04 '18 at 12:37

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For $p\ge 7$, this occurs iff the quadratic field $\Bbb Q(\sqrt{-p})$ has class number one. This is a consequence of the analytic class number formula. There are only a finite number of such fields. The largest $p$ for which this is true is $p=163$.

answered Mar 04 '18 at 12:37

Angina Seng

161,540

can you give this reslut some paper or links?Thanks – math110 Mar 04 '18 at 12:47
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@communnites https://math.stackexchange.com/questions/533338/sum-of-quadratic-residues/549263 – Angina Seng Mar 04 '18 at 12:55
I can't understand this why 163 is largest ? – math110 Mar 04 '18 at 13:49
I don't see how the fractional part of $k^2/p$ relates to the Legendre symbol... ? – Alphonse Mar 04 '18 at 21:40
This answer seems wrong to me (as I mentioned already in my comments six months ago). The question is about $\left{\dfrac{k^2}{p}\right}$ where ${x}=x-[x]$, so I don't see any link with quadratic residues or class numbers. – Alphonse Nov 22 '18 at 15:24

Is there finite number of $p$ such $\sum_{k=1}^{p-1}\left\{\frac{k^2}{p}\right\}=\frac{p-3}{2}？$

1 Answers1