Prove $4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}$ 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!
 A: This goes way back to Emma Lehmer, see her elementary paper on Fermat quotients and Bernoulli numbers from 1938.
Assume $p\ge 7$. First, I reformulate your congruence. I replace
$$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}$$
with the equivalent
$$\sum_{k=1}^{p-1} \frac{1}{k^2} \equiv 8\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{(p-2k)^2}\pmod{p^3}$$
Lehmer has proved the following, see equation (19) in the linked paper:
$$\sum_{k=1}^{\frac{p-1}{2}} (p-2k)^{2r} \equiv p 2^{2r-1} B_{2r} \pmod {p^3}$$
(valid when $2r \not\equiv 2 \pmod {p-1}$). Plugging $2r=p^3-p^2-2$, we get
$$\sum_{k=1}^{\frac{p-1}{2}} (p-2k)^{p^3-p^2-2} \equiv p 2^{p^3-p^2-3} B_{p^3-p^2-2} \pmod {p^3}$$
Now use Euler's theorem to replace $x^{p^3-p^2-2}$ with $x^{-2}$:
$$\sum_{r=1}^{\frac{p-1}{2}} \left(\frac{1}{p-2k}\right)^2 \equiv p 2^{-3} B_{p^3-p^2-2} \pmod {p^3}$$
So your congruence becomes
$$(*)\qquad \sum_{k=1}^{p-1} \frac{1}{k^2} \equiv pB_{p^3-p^2-2} \pmod {p^3}$$
This was also proved in Lehmer's paper - in equation (15) she writes
$$\sum_{k=1}^{p-1} k^{2r} \equiv pB_{2r} \pmod {p^3}$$
(valid when $2r \neq 2 \mod {p-1}$). Plugging $2r=p^3-p^2-2$ and using Euler's theorem, we get the desired result.

It is now easy to generalize - when $2r \not\equiv 2 \pmod {p-1}$, we have
$$\sum_{k=1}^{p-1} (-1)^k k^{2r} \equiv (1-2^{2r}) \sum_{k=1}^{p-1} k^{2r} \pmod {p^3}$$
Plugging $2r=p^3-p^2-2c$ ($c \not\equiv -1 \pmod {\frac{p-1}{2}}$) we get
$$\sum_{k=1}^{p-1} \frac{(-1)^k}{k^{2c}} \equiv (1-2^{-2c}) \sum_{k=1}^{p-1} \frac{1}{k^{2c}} \pmod {p^3}$$

To get a taste of things, I include a proof of $(*)$. Via Faulhaber's formula for sum of powers, we get:
$$\sum_{k=1}^{p-1} \frac{1}{k^2} \equiv \sum_{k=1}^{p-1} k^{p^3-p^2-2} \equiv \sum_{k=1}^{p} k^{p^3-p^2-2} = $$
$$\frac{1}{p^3-p^2-1} \sum_{j=0}^{p^3-p^2-2}\binom{p^3-p^2-1}{j}B_j p^{p^3-p^2-1-j} \pmod {p^3}$$
Note that odd-indexed Bernoulli's vanish, and that by the Von Staudt–Clausen theorem, the even-indexed Bernoulli's have $p$ in the denominator with multiplicity $\le 1$. Additionally, $B_{p^3-p^2-4}$ is $p$-integral (since $p-1 \nmid p^3-p^2-4$).
Hence, for the purpose of evaluating the sum modulo $p^3$,  we can drop almost all the terms and remain only with the last one:
$$\sum_{k=1}^{p-1} \frac{1}{k^2} = \frac{1}{p^3-p^2-1} \binom{p^3-p^2-1}{p^3-p^2-2} B_{p^3-p^2-2} p  = pB_{p^3-p^2-2} \pmod {p^3}$$
