Prove that the following congruence involving Lucas sequence is true

Question

I am proving a congruence that appears in a paper, it claims that for $t\in\mathbb{Z}$ and prime $p\geq5$ $$p\sum_{k=1}^{p-1} \frac{t^k}{k^2\binom{2k}{k}}\equiv \frac{2-v_p(2-t)-t^p}{2p}-p^2\sum_{k=1}^{p-1} \frac{v_k(2-t)}{k^3} \pmod{p^3}$$ where $v_n(P)$ is the Lucas sequence when $Q=1$, i.e. $v_n(P)=a^n+b^n$ where $a$ and $b$ are the roots of $x^2-Px+1=0$. The paper uses the following identity \begin{align*}\begin{split}p\binom{2p}{p}\sum_{k=1}^{p-1} \frac{t^k}{k^2\binom{2k}{k}} &= \frac{v_p(t-2)-t^p}{p}+p\sum_{k=1}^{p-1} \binom{2p}{p-k}\frac{v_k(t-2)}{k^2}+p\binom{2p}{p}H_{p-1}^{(2)}+\frac{1}{p}\binom{2p}{p}\\ &= \frac{v_p(t-2)-t^p}{p}+p\sum_{k=1}^{p-1} \binom{2p}{k}\frac{v_{p-k}(t-2)}{(p-k)^2}+p\binom{2p}{p}H_{p-1}^{(2)}+\frac{1}{p}\binom{2p}{p}\end{split}\end{align*} Now using the fact that $\binom{2p}{p}\equiv 2\pmod{p^3}$ and $p\binom{2p}{k}\equiv \frac{2p^2}{k}(-1)^{k-1} \pmod{p^3}$ for $k=1,\dots,p-1$. It is then easy to deduce that $$p\binom{2p}{p}\sum_{k=1}^{p-1} \frac{t^k}{k^2\binom{2k}{k}}\equiv 2p\sum_{k=1}^{p-1} \frac{t^k}{k^2\binom{2k}{k}} \pmod{p^3}$$ and \begin{align*}\begin{split}p\sum_{k=1}^{p-1} \binom{2p}{k}\frac{v_{p-k}(t-2)}{(p-k)^2} &\equiv 2p^2\sum_{k=1}^{p-1} \frac{(-1)^{k-1}v_{p-k}(t-2)}{k(p-k)^2} \pmod{p^3}\\ &\equiv 2p^2\sum_{k=1}^{p-1} \frac{(-1)^{p-k-1}v_{k}(t-2)}{(p-k)k^2} \pmod{p^3}\\ &\equiv 2p^2\sum_{k=1}^{p-1} \frac{v_{k}(2-t)}{(p-k)k^2} \pmod{p^3}\\ &\equiv -2p^2\sum_{k=1}^{p-1} \frac{v_k(2-t)}{k^3} \pmod{p^3}\end{split}\end{align*} since we have $v_k(-x)=(-1)^k v_k(x)$. Now the problem arise from the last two expression, notice that we have $$p\binom{2p}{p}H_{p-1}^{(2)}=p\Bigg(\binom{2p}{p}-2+2\Bigg)H_{p-1}^{(2)}\equiv 2pH_{p-1}^{(2)} \pmod{p^3}$$ Hence it suffices to prove that $$2pH_{p-1}^{(2)}+\frac{1}{p}\binom{2p}{p}\equiv \frac{2}{p}\pmod{p^3}$$ which is equivalent to prove that $$2p^2H_{p-1}^{(2)}\equiv 2-\binom{2p}{p}\pmod{p^4}$$ and I could not prove this, the paper says using Wolstenholme's theorem $H_{p-1}^{(2)}\equiv 0 \pmod{p}$, yet I have no idea how could I use this, any helps would be appreciated.

Answer 1

First note that $$ \tag{1} 2-{2p \choose p}= 2\Big(1-{2p-1 \choose p-1}\Big).$$

Also from the well-known relations between the roots and coefficients of a polynomial, we have $$ \tag{2} {X-1 \choose j}=\sum_{k=0}^j (-1)^{j-k}G_j^{(k)}X^k,$$ where $$G_j^{(k)}=\sum_{0\le i_1 \lt i_2 \cdot\cdot \lt i_k \le j}\frac{1}{i_1i_2 \cdot\cdot i_k} \text{ and }G_j^{(0)}=1$$ and from the well-known relation between elementary symmetric polynomials and sums of powers, we have $$ \tag{3} G_n^{(k)}=-\frac{1}{k}\sum_{j=1}^k(-1)^jH_n^{(j)}G_n^{(k-j)}$$ which gives $$\tag{4} G_n^{(1)}=H_n$$ $$\tag{5} G_n^{(2)}=\frac{1}{2}\Big(H_n^2-H_n^{(2)}\Big)$$ $$\tag{6} G_n^{(3)}=\frac{1}{6}\Big(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\Big)$$ Then with $X=2p$ and $j=p-1$ Equation (2) becomes $$ {2p-1 \choose p-1}=\sum_{k=0}^{p-1} (-1)^{k}G_{p-1}^{(k)}2^kp^k,$$ and since the denominator in $G_{p-1}^{(k)}$ is coprime to $p$ we have: $$ {2p-1 \choose p-1}\equiv G_{p-1}^{(0)}-G_{p-1}^{(1)}2p+ G_{p-1}^{(2)}4p^2 -G_{p-1}^{(3)}8p^3 \pmod {p^4}$$ and accounting for (4), (5) and (6) we have $${2p-1 \choose p-1}\equiv 1-2pH_{p-1}+ 2p^2\big(H_{p-1}^2-H_{p-1}^{(2)}\big) -\frac{4p^3}{3}\Big(H_{p-1}^3-3H_{p-1}H_{p-1}^{(2)}+2H_{p-1}^{(3)}\Big) \pmod {p^4}$$ It is also well-known that for any $k\ge 0$, we have $H_{p-1}^{2k+1} \equiv 0 \pmod {p}$, you see this by pairing the 2 terms symetric around the middle of the harmonic sum, so that since $p \gt3$, the last term on the rhs of the latter congruence disappears. The term $H_{p-1}^2$ in $p^2$ also disappears and we are left with

$$\tag{**}{2p-1 \choose p-1}\equiv 1-2p\Big(H_{p-1}+pH_{p-1}^{(2)}\Big) \pmod {p^4}.$$

Now $$ H_{p-1}=\sum_{i=1}^{p-1}\frac{1}{i}=\sum_{i=1}^{p-1}\frac{1}{(p-i)}=-\sum_{i=1}^{p-1}\frac{1}{i(1-\frac{p}{i})}$$ $$ H_{p-1}=-\sum_{i=1}^{p-1}\sum_{j\ge 0}\frac{p^j}{i^{j+1}}= -\sum_{j\ge 0}p^j\sum_{i=1}^{p-1}\frac{1}{i^{j+1}}=-\sum_{j\ge 0}p^jH_{p-1}^{(j+1)}$$

$$ 2H_{p-1}=-\sum_{j\ge 1}p^jH_{p-1}^{(j+1)}= -pH_{p-1}^{(2)}-p^2H_{p-1}^{(3)}-p^3H_{p-1}^{(4)} - \cdot\cdot $$ $$2H_{p-1}+2pH_{p-1}^{(2)} =pH_{p-1}^{(2)}-p^2H_{p-1}^{(3)}-p^3H_{p-1}^{(4)} - \cdot\cdot $$ $$2p\Big(H_{p-1}+pH_{p-1}^{(2)}\Big)= p^2H_{p-1}^{(2)}-p^3H_{p-1}^{(3)} - \cdot\cdot $$ then $$2p\Big(H_{p-1}+pH_{p-1}^{(2)}\Big) \equiv p^2H_{p-1}^{(2)} \pmod {p^4} $$ then (**) becomes $$\tag{***}{2p-1 \choose p-1}\equiv 1-p^2H_{p-1}^{(2)} \pmod {p^4}$$ which together with (1) gives

$$2p^2H_{p-1}^{(2)}\equiv 2-\binom{2p}{p}\pmod{p^4}$$ provided that $p \ge 5$.