"Wolstenholme prime" Related question. $Q$.
If $p$ is an odd prime, show that
$$
\left(\begin{array}{c}
2 p-1 \\
p-1
\end{array}\right) \equiv 1 \bmod p^{2}
$$
My approach
Here,
$$
\begin{array}{l}
\left(\begin{array}{c}
2 p-1 \\
p-1
\end{array}\right)=\frac{(2 p-1) !}{(p-1) !(2 p-1-p+1) !} \\
=\frac{(2 p-1) !}{(p-1) !(p) !}
\end{array}
$$
How it can be show that this is congruent to $1(mod $p$²$$)$?
And also tell me that is there any relation with wolstenholme prime?
 A: So let $\frac{(2p-1)!}{(p-1)!p!}=a+kp^2$ for some naturals $a,k$, $0\leq a\leq p^2-1$.
Simplifying a bit, this implies
\begin{equation}
(p+1)\dots(2p-1)=(p-1)!(a+kp^2).
\end{equation}
Now Wilson's theorem tells us that $(p-1)!\equiv-1 \mod{p}$ and this immediately implies that $(p+1)\dots(2p-1)\equiv -1\mod{p}$ as well. So let $m,n\in\mathbb{N}$ be such that $(p-1)!=-1+mp$ and $(p+1)\dots(2p-1)=-1+np$.
Plugging back into our equation this means $-1+np=(-1+mp)(a+kp^2)$ which implies $p|a-1$, so we can write $a=pb+1$ for some $0\leq b\leq p-1$.
Then our equation becomes $(p+1)(p+2)\dots(p+(p-1))=(p-1)!(1+p(b+kp))$ implying, by opening up the factors on the LHS, that for some natural $\ell$ we have
\begin{equation*}
p(\sum_{j=1}^{p-1}\frac{(p-1)!}{j})+\ell p^2=pb+kp^2.
\end{equation*}
Therefore $b\equiv \sum_{j=1}^{p-1}\frac{(p-1)!}{j}\mod{p}$, so we are left to show that $p|S$, where $S:=\sum_{j=1}^{p-1}\frac{(p-1)!}{j}$.
Observe that based again on Wilson's theorem, $S\equiv\sum_{j=1}^{p-1}(-1)j^{-1}\mod{p}$, where by $j^{-1}$ we mean the inverse of $j \mod{p}$. That means $S\equiv(-1)(1^{-1}+(p-1)^{-1})-\sum_{j=2}^{p-2}j^{-1}\equiv -\sum_{j=2}^{p-2}j^{-1}\equiv-\sum_{j=2}^{p-2}j\equiv 0$, where in the last step we can pair the terms up (i.e., $j$ with $p-j$) noting that there are even many summands since $p$ was odd.
Not familiar with Wolstenholme primes, but apparently his theorem states that the desired equivalence holds $\mod{p^3}$ as well for all $p>3$ and a Wolstenholme prime is a prime for which the equivalence holds $\mod{p^4}$. (See here https://en.wikipedia.org/wiki/Wolstenholme_prime)
A: We have $$\left(\begin{array}{c}
2 p-1 \\
p-1
\end{array}\right)=\frac{(p+1)(p+2)\cdots(p+p-1)}{(p-1)!}=\frac{Mp^2+Ap+(p-1)!}{(p-1)!}$$ What the value modulo p of $A$ is? It is zero.
In fact, by Vieta's coefficients, $A$ is equal to
$$\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\cdots+\frac{(p-1)!}{p-1}=(p-1)!\left(1+\frac12+\frac13+\cdots\frac{1}{p-1}\right)$$ Now the last factor corresponds bijectively with the inverses in the field $\mathbb F_p$ so is equal to
$1+2+3\cdots+(p-1)$ which is equal to $0$ modulo $p$ then $Ap$ is equal to $0$ modulo $p^2$.
Thus $$\frac{Mp^2+Ap+(p-1)!}{(p-1)!}\equiv\frac{(p-1)!}{(p-1)!}=1\pmod{p^2}$$ This finish the proof.
