TITLE: How to prove $\sum_{k=1}^{p-1}\frac{2^k}{k^2}\equiv-\frac{(2^{p-1}-1)^2}{p^2}\pmod p$
QUESTION [17 upvotes]: Let prime number $p\ge 5$,show that
$$\left(\dfrac{2}{1^2}+\dfrac{2^2}{2^2}+\cdots+\dfrac{2^{p-1}}{(p-1)^2}\right)\equiv -\left(\dfrac{2^{p-1}-1}{p}\right)^2\pmod p$$
I want use  Wolstenholme's theorem
$$\sum_{k=1}^{p-1}\dfrac{1}{k}\equiv 0\pmod {p^2}$$
and
$$\binom{p-1}{i}\equiv (-1)^i\pmod p$$
But failed

REPLY [15 votes]: There must be a shorter way to do this.
Claim 1: For $p\geq 3$,
$$
\frac{2^{p-1}-1}{p}\equiv -\frac{1}{2}\sum_{n=1}^{p-1}\frac{(-1)^n}{n}+\frac{p}{2}\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^n}{n\cdot m}\mod p^2.
$$
Proof:
\begin{align*}
\frac{2^{p-1}-1}{p}&=\frac{1}{2p}\sum_{n=1}^{p-1}{p\choose n}\\
&=\frac{1}{2p}\sum_{n=1}^{p-1}\frac{p}{n}\left
(\frac{p-1}{1}\cdots\frac{p-(n-1)}{n-1}\right)\\
&= -\frac{1}{2}\sum_{n=1}^{p-1}\frac{(-1)
^n}{n}\left(1-\frac{p}{1}\right)\cdots \left(1-\frac{p}{n-1}\right)\\
&\equiv -\frac{1}{2}\sum_{n=1}^{p-1}\frac{(-1)^n}{n}+\frac{p}{2}\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^n}{n\cdot m}\mod p^2.
\end{align*}
Claim 2: For $p\geq 3$,
$$
\left(\frac{2^{p-1}-1}{p}\right)^2\equiv \sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^{n+m}}{n\cdot m}\mod p.
$$
Proof: Reduce Claim 1 modulo $p$ and square both sides to obtain
\begin{align*}
\left(\frac{2^{p-1}-1}{p}\right)^2&\equiv \frac{1}{4}\sum_{n,m=1}^{p-1}\frac{(-1)^{n+m}}{n\cdot m}\mod p\\
&=\frac{1}{2}\sum_{p-1\geq n>m\geq 1}\frac{(-1)^{n+m}}{n\cdot m}+\frac{1}{4}\sum_{n=1}^{p-1}\frac{1}{n^2}\\
&\equiv \frac{1}{2}\sum_{p-1\geq n>m\geq 1}\frac{(-1)^{n+m}}{n\cdot m}\mod p.\\
\end{align*}
Claim 3: For $p\geq 5$,
$$
\sum_{n=1}^{p-1}\frac{2^n}{n^2}\equiv -\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^n}{n\cdot m}\mod p.
$$
First, observe that
\begin{align*}
1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{p}&=
\frac{1}{2}\int_{-1}^1\left(1+x+x^2+\ldots+x^{p-
1}\right)\,dx\\
&=\frac{1}{2}\int_{-1}^1\frac{x^p-1}{x-1}\,dx\\
&=\frac{1}{2}\int_0^2\frac{(1-x)^p-1}{x}\,dx\\
&=\frac{1}{2}\sum_{n=1}^p {p\choose n}\frac{(-2)^n}{n}\\
&=\frac{2^{p-1}}{p}+\frac{1}{2}\sum_{n=1}^{p-1}\frac{p\cdot (-2)^n}{n^2} \left
(\frac{p-1}{1}\cdots\frac{p-(n-1)}{n-1}\right).\\
\end{align*}
It follows that
\begin{gather*}
1+\frac{1}{3}+\ldots+\frac{1}{p-2}\equiv \frac{2^{p-1}-1}{p}-\frac{p}{2}\sum_{n=1}^{p-1}\frac{2^n}{n^2}\mod p^2,\\
\frac{1}{2}\left(\sum_{n=1}^{p-1}\frac{1}{n}-\sum_{n=1}^{p-1}\frac{(-1)^n}{n}\right)\equiv -\frac{1}{2}\sum_{n=1}^{p-1}\frac{(-1)^n}{n}+\frac{p}{2}\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^n}{n\cdot m}-\frac{p}{2}\sum_{n=1}^{p-1}\frac{2^n}{n^2}\mod p^2,\\
\sum_{n=1}^{p-1}\frac{2^n}{n^2}\equiv \sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^n}{n\cdot m}\mod p.
\end{gather*}
We used Wolstenholme's congruence on the last line.
Claim 4: For $p\geq 3$,
$$
\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^{n+m}}{n\cdot m}\equiv 2\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^{n}}{n\cdot m}\mod p.
$$
Proof: Let $A_p$ be the coefficient of $x^{p-1}$ in the power series expansion of
$$
\frac{\log(1+x)^2}{1-x}.
$$
We compute $A_p$ in two different ways. First:
\begin{align*}
A_p&=\sum_{m=1}^{p-1}\sum_{n=1}^{p-1-m}\frac{(-1)^{n+m}}{n\cdot m}\\
&=-\sum_{m=1}^{p-1}\sum_{n=m+1}^{p-1}\frac{(-1)^{n+m}}{(p-n)\cdot m}\\
&\equiv \sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^{n+m}}{n\cdot m}\mod p.
\end{align*}
Second, we have
\begin{align*}
\frac{\log(1+x)^2}{1-x}&=\frac{1}{1-x}\left(\int_0^x\frac{dt}{1+t}\right)^2\\
&=\frac{2}{1-x}\int_0^x\left(\int_0^y\frac{dt}{1+t}\right)\frac{dy}{1+y}\\
&=\frac{2}{1-x}\int_0^x\frac{\sum_{m\geq 1}(-1)^{m-1}\frac{y^m}{m}}{1+y}dy\\
&=\frac{2}{1-x}\int_0^x\left(\sum_{n=1}^{\infty}\sum_{m=1}^{n}\frac{(-1)^{n-1}}{m} y^n\right)dy\\
&=\frac{2}{1-x}\sum_{n=1}^{\infty} \sum_{m=1}^n\frac{(-1)^{n-1}}{m\cdot(n+1)}x^{n+1}.\\
\end{align*}
It follows that
$$
A_p=2\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^{n}}{n\cdot m}.
$$
Proof of the desired congruence:
\begin{align*}
-\left(\frac{2^{p-1}-1}{p}\right)^2&\equiv -\frac{1}{2}\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^{n+m}}{n\cdot m}\\
&\equiv-\sum_{n=1}^{p-1}\sum_{m=1}^{n-1}\frac{(-1)^{n}}{n\cdot m}\\
&\equiv \sum_{n=1}^{p-1}\frac{2^n}{n^2}\mod p,
\end{align*}
where the first line is Claim 2, the second line is Claim 4, and the third line is Claim 3.