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let $p=8k+1$ prime,prove or disprove $$\sum_{k=1}^{p-1}\dfrac{(-1)^k}{k}\left(\dfrac{k}{p}\right)\equiv 0\pmod {p^2}\tag{1}$$ where $\left(\dfrac{k}{p}\right)$ is Legendre symbol

It seem this Notes on Wolstenholme’s Theorem How prove this $(p-1)!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1}\right)\equiv 0\pmod{p^2}$ I can't prove $(1)$

JMP
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  • It is $=\sum_{k= (p+1)/2}^{p-1} \frac1k(\frac{k}p)$. Why is it $\equiv 0\bmod p$, what is the point of $(-1)^k$ ? – reuns Nov 20 '20 at 04:46
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    Indeed: as it is stated right now, it is false for $p = 17$. Very false:

    $$\sum_{k=1}^{16}\frac{(-1)^k}{k}\left(\frac{k}{17}\right) = \frac{20519}{720720} $$

     – Good Boy Nov 20 '20 at 17:50
  • @GoodBoy: Everything's fine modulo $p^2$ ($20519=71\cdot17^2$, and $720720$ is coprime to $17$). – metamorphy Nov 22 '20 at 12:43

1 Answers1

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This is what I thought of. It is not finished but it is a start.

Observe that $$\frac{1}{p}\binom{p}{k}=\frac{1}{p}\cdot\frac{p!}{k!\cdot(p-k)!}\equiv\frac{(p-1)!}{k!\cdot-k\cdot...\cdot-(p-1)}\equiv\frac{(p-1)!}{(p-1)!\cdot k}\cdot(-1)^{p-k}\equiv\frac{(-1)^{k+1}}{k}\pmod{p}$$

So $$\frac{1}{p}\binom{p}{k}\equiv\frac{(-1)^{k+1}}{k}\pmod{p}\Rightarrow\binom{p}{k}\equiv\frac{p\cdot(-1)^{k+1}}{k}\pmod{p^2}$$

To prove $$\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\bigg(\frac{k}{p}\bigg)$$ is divisible by $p^2$, it is enough to prove that $p^2$ divides $$\frac{1}{p}\bigg(\sum_{k=1}^{p-1}\frac{p\cdot(-1)^k}{k}\bigg(\frac{k}{p}\bigg)\bigg)\equiv\frac{1}{p}\bigg(\sum_{k=1}^{p-1}\binom{p}{k}\cdot\bigg(\frac{k}{p}\bigg)\bigg)\pmod{p^2}$$ But $$\sum_{k=1}^{p-1}\binom{p}{k}\cdot\bigg(\frac{k}{p}\bigg)=2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-\sum_{k=1}^{p-1}\binom{p}{k}=2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-2^p+2$$ so we want to prove that $$\frac{1}{p}\bigg(2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-2^p+2\bigg)$$ is divible by $p^2$, in other words $$\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}+1\equiv2^{p-1}\pmod{p^3}$$