Dgdrxrt

Show that
$$(p-1)!left(1+dfrac12+dfrac13+cdots+dfrac1p-1right)equiv 0pmodp^2.$$

Maybe use this
$$dfrac1k+dfrac1p-k=dfracpk(p-k)$$
and then I can't. Can you help me to prove it?

Thank you.

The solution below is adapted from Notes on Wolstenholme’s Theorem by Timothy H. Choi.

Let
$$
S=(p-1)!sum_k=1^p-1 frac1k
$$
Using your insight
$$
dfrac1k+dfrac1p-k=dfracpk(p-k)
$$
we have
$$
2S=(p-1)!sum_k=1^p-1 left(dfrac1k+dfrac1p-kright) =
psum_k=1^p-1 frac(p-1)!k(p-k) = pS'
$$
Note that $S'$ is an integer. Now
$$
frac(p-1)!k(p-k) equiv (k^2)^-1 bmod p
$$
where the inverse is taken $bmod p$. This is a consequence of Wilson’s Theorem.
Hence
$$
S'equiv
sum_k=1^p-1 (k^2)^-1 equiv
sum_k=1^p-1 k^2 = frac(p-1)p(2(p-1)+1)6 equiv 0 bmod p
$$
This means that $2Sequiv 0 bmod p^2$ and so $Sequiv 0 bmod p^2$. (We need $p>3$ twice here.)

As others have noted, the congruence is not true for $p=3$, since
$$ 2!left(1+frac 1 2right)=2+1=3,$$
which is not divisible by $9$. We can still use what you suggested to prove the congruence holds $operatornamemod p$. Let $p$ be an odd prime, then
beginalign*
(p-1)!sum_k=1^p-1 frac 1 k &= (p-1)! sum_k=1^(p-1)/2 left(frac 1 k + frac 1 p-kright) \&= (p-1)!sum_k=1^(p-1)/2 fracpk(p-k) = psum_k=1^(p-1)/2 frac(p-1)!k(p-k).
endalign*
Since $(p-1)!$ always contains $k$ and $(p-k)$ as a factor, the fractions in the sum are integers and the result is a multiple of $p$.

See lhf's answer for why it is even a multiple of $p^2$ as long as $p>3$.

I came up with this proof while I work on this problem.

We write $f(x) equiv_p g(x)$ if polynomials $f(x), g(x) in mathbbZ[x]$ satisfies $a_iequiv b_i$ mod $p$ for all coefficients $a_i$ of $f(x)$, and $b_i$ of $g(x)$.

Then we have by Fermat's theorem,
$$
x^p-1-1 equiv_p (x-1)(x-2) cdots (x-(p-1)).
$$

Group the numbers in pairs as discussed here already,
$$
frac 11+frac 1p-1=frac p1(p-1), frac12+ frac 1p-2=frac p2(p-2), cdots, frac 1(p-1)/2+frac1(p+1)/2=frac p (p-1)(p+1)/4 .
$$
Each group has numerator divisible by $p$. Then it suffices to prove that the numerator $N$ in
$$
sum_j=1^(p-1)/2 frac 1j(p-j)=frac N(p-1)! (*)
$$
is divisible by $p$.

We may write

beginalign*x^p-1-1&equiv_p
(x-1)(x-2)cdots (x-(p-1))\ &= prod_j=1^(p-1)/2 (x-j)(x-(p-j)) \ &=prod_j=1^(p-1)/2 (x^2-(j+(p-j))x + j(p-j))\
&equiv_pprod_j=1^(p-1)/2 (x^2+j(p-j))
endalign*
Expanding the last product, we find that $N$ is the coefficient of $x^2$. Since $pgeq 5$, the coefficient of $x^2$ in the product must be divisible by $p$. Hence, $p|N$ as desired.