Recently I think on the following conjecture of binomial coefficient:
There are no positive integers $a>b$ such that
$\binom{an}{bn}\equiv 0\pmod{pn-1}$ for all $n\geqslant 1$.
I could prove this conjecture for the special cause $\gcd(a,p)=1$. I proved the following theorem:
Theorem: There are no positive integers $a>b$ with $\gcd(a,p)=1$ such that
$\binom{an}{bn}\equiv 0\pmod{pn-1}$ for all $n\geqslant 1$.
Proof:Suppose on the contrary that $a>b$ exist satisfying the conditions of the theorem. Let $a=a'p-a_0$ and $b=b'p+b_0$ such that $0\leq a_0,b_0<p$, and $a',b'\geq0$. Clearly, $a+a'-1\geq b+b'$. Let $n\geq 2a$ be such that $q_n:=pn-1$ is a prime (whose existence is guaranteed by Dirichlet's theorem). If $m_n:=pn+n-1$, then $q_n(p+1)=pm_n-1$ divides $\binom{am_n}{bm_n}$. On the other hand, \begin{align*} \binom{am_n}{bm_n}&=\binom{a(pn+n-1)}{b(pn+n-1)}\\ &=\binom{(a+a'-1)q_n+nq_n-a_0n+a'}{(b+b')q_n+b_0n+b'}\\ &\equiv\binom{a+a'-1}{b+b'}\binom{nq_n-a_0n+a'}{b_0n+b'}\\ &\equiv\binom{a+a'-1}{b+b'}\binom{-a_0n+a'}{b_0n+b'}\not\equiv0\pmod{q_n} \end{align*} by Lucas's theorem, which is a contradiction.
I think prove of this conjecture for generl case is very hard and actully I have no idea for prove it.
I would be happy if I know your comments.
Thanks in advance.