show that $p|\sum_{j=0}^{n}\binom{n}{j}^4$ 
Let $p$ be prime number,and postive integer $n$ such $$3n<3p<4n$$
  show that
  $$p|\sum_{j=0}^{n}\binom{n}{j}^k~~~~,k=4$$

I have Solve when $k=2$.because use indentity 
$$\sum_{j=0}^{n}\binom{n}{j}^2=\binom{2n}{n}$$  since $n<p<\dfrac{4}{3}n$  then use Lucas thereom we have
$$\binom{2n}{n}\equiv 0\pmod p$$
so I have prove 
$$p|\sum_{j=0}^{n}\binom{n}{j}^k~~~~,k=2$$
But for $k=4$,it seem hard to prove it。Thanks 
 A: Let 
$$a(n)=\sum_{j=0}^{n}\binom{n}{j}^4$$
then according to A005260 it satisfies the recurrence
$$n^3a(n) = 2(2n - 1)(3n^2 - 3n + 1)a(n-1) + (4n - 3)(4n - 4)(4n - 5)a(n-2)\tag{1}.$$
Let $p$ be a prime then
$$a(p-1)=\sum_{j=0}^{p-1}\binom{p-1}{j}^4\equiv \sum_{j=0}^{p-1}((-1)^j)^4=p\equiv 0\pmod{p}.$$
Note that by the recurrence (1)
$$0\equiv p^3a(p) = 2(2p - 1)(3p^2 - 3p + 1)a(p-1) + (4p - 3)(4p - 4)(4p - 5)a(p-2)\\\equiv -2a(p-1)-60 a(p-2)\equiv -60 a(p-2)$$
which implies that for $p>5$, $a(p-2)\equiv 0$ modulo $p$.
Now let $n$ be an integer number $n<p-2$ and such that $3n<3p<4n$, that is $(3p/4)<n<p-2$, then $p$ does not divide
$$(4(n+2) - 3)(4(n+2) - 4)(4(n+2) - 5)=(4n+5)(4n+4)(4n+3)$$
and by using the recurrence (1) we show that $a(n)\equiv 0$ modulo $p$.
A: I will quote the solution from here.

Let $p = n+m$. Then one has
  $$\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{(1)(2)...(k)} \equiv (-1)^k \frac{(m)(m+1)...(m+k-1)}{(1)(2)...(k)} \equiv (-1)^k \frac {(k+1)(k+2)...(m+k-1)}{(1)(2)...(m-1)} \pmod p$$
  Now taking the fourth power and clearing denominators, our sum becomes
  $$ \sum_{i=0}^{n} ((i+1)(i+2)...(m+i-1))^4 \equiv \sum_{i=0}^{p-1} ((i+1)(i+2)...(m+i-1))^4 \pmod p$$
  Note that our sum is of the form $\sum_{i=0}^{p-1} P(i) \pmod {p}$ where $P(x) = ((x+1)(x+2)...(x+m-1))^4$ and it is well known that if the degree of $P$ is $\leq p-2$ then the sum is $\equiv 0 \pmod p$. But the degree of our polynomial is $4(p-n-1)$ which is always $< p-1$ as $3p < 4n$. Hence we are done.

