We are given $2017$ prime numbers $p_1,p_2,\ldots,p_{2017}$. Prove that $\displaystyle \prod_{i
Note that $5777 = 53 \cdot 109$. We first consider $p_i^{p_j}-p_j^{p_i}$ modulo $53$. We are to prove that \begin{align*}p_i^{p_j} \equiv p_j^{p_i} \pmod{53} \quad \tag{1}\end{align*} for some primes $p_i,p_j$.There exist primes in the list such that $p_i = 53k_1+d,p_j = 53k_2+d$, where $0 \leq d \leq 52$ and $k_1,k_2 > 0$. Then $(1)$ is equivalent to $$d^{53k_1} \equiv d^{53k_2} \pmod{53} \iff d^{53(k_1-k_2)} \equiv 1 \pmod{53}.$$ Now since $53$ is odd, $k_1,k_2$ must be only odd or only even. Then note there are at least $2017-15 = 2002$ such primes $p_i,p_j$ because there are $28$ primes between $0$ and $53$. Therefore, since $\left\lceil\dfrac{2002}{53}\right\rceil = 39$, there exist $p_i,p_j$ such that $k_1 \equiv k_2 \pmod{26}$ and $k_1$ and $k_2$ have the same parity quotient upon division by $26$ and $k_1 > k_2$. Therefore, $$52 \mid 53(k_1-k_2)$$ and so we have found primes $p_i,p_j$ that satisfy $(1)$.
I tried proving divisibility by $109$ using the same argument, but it didn't work:
Now we consider modulo $109$. We are to prove that \begin{align*}p_i^{p_j} \equiv p_j^{p_i} \pmod{109} \quad \tag{2}\end{align*} for some primes $p_i,p_j$. primes in the list such that $p_i = 109m_1+d_1,p_j = 109m_2+d_1$, where $0 \leq d_1 \leq 108$ and $m_1,m_2 > 0$. Then $(2)$ is equivalent to $$d_1^{109m_1} \equiv d_1^{109m_2} \pmod{109} \iff d_1^{109(m_1-m_2)} \equiv 1 \pmod{109}.$$ Now since $109$ is odd, $m_1,m_2$ must be only odd or only even. Then note there are at least $2017-28 = 1989$ such primes $p_i,p_j$ because there are $28$ primes between $0$ and $109$. Therefore, $\left\lceil\dfrac{1989}{109}\right\rceil = 19$.
Is it possible to use the same argument to prove divisibility by $109$?