prove or disprove $\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\left(\frac{k}{p}\right)\equiv 0\pmod{ p^2}$ 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 $(1)$
 A: 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 $$\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}+1\equiv2^{p-1}\pmod{p^3}$$
