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Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

Comments:

(1) Since $$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$ we need to prove $$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$ This idea was given by Fedor Petrov but there are something wrong in his answer. This question is still open.

(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e., $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}. $$ This was conjectured by tkr.

I appreciate any proofs, hints, or references!

Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

Comments:

(1) Since $$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$ we need to prove $$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$ This idea was given by Fedor Petrov.

(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e., $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}. $$ This was conjectured by tkr.

I appreciate any proofs, hints, or references!

Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

Comments:

(1) Since $$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$ we need to prove $$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$ This idea was given by Fedor Petrov but there are something wrong in his answer. This question is still open.

(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e., $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}. $$ This was conjectured by tkr.

I appreciate any proofs, hints, or references!

Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

Comments:

(1) Since $$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$ we need to prove $$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$ This idea was given by Fedor Petrov but there are something wrong in his answer. This question is still open.

(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e., $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}. $$ This was conjectured by tkr.

I appreciate any proofs, hints, or references!

Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

Comments: Since $$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$ we need to prove $$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$ This idea was given by Fedor Petrov but there are something wrong in his answer. This question is still open.

I appreciate any proofs, hints, or references!

Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

Comments:

(1) Since $$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$ we need to prove $$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$ This idea was given by Fedor Petrov but there are something wrong in his answer. This question is still open.

(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e., $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}. $$ This was conjectured by tkr.

I appreciate any proofs, hints, or references!