$n$ is prime iff $\binom{n^2}{n} \equiv n \pmod{n^4}$? Can you prove or disprove the following claim:

Let $n$ be a natural number greater than two , then $$n \text{ is prime iff } \binom{n^2}{n} \equiv n \pmod{n^4}$$

You can run this test here. I have verified this claim for all $n$ up to $100000$ .

Note that $\displaystyle\binom{n^2}{n} = \frac{1}{(n - 1)!} \frac{n^2 (n^2 - 1) ... (n^2 - (n - 1))}{n} = \frac{1}{(n - 1)!} n (n^2 - 1) ... (n^2 - (n - 1))$
Consider a prime $p > 2$. Then $1, 2, ..., p - 1$ are all invertible modulo $p^4$; thus, so is $(p - 1)!$.
Now consider $\displaystyle\binom{p^2}{p} = \frac{1}{(p - 1)!} p (p^2 - 1) ... (p^2 - (p - 1))$.
Define the polynomial $P(x) = x (x^2 - 1) (x^2 - 2) ... (x^2 - (p - 1))$. We wish to reduce $P(x)$ modulo $x^4$. We note that this will only have an $x$ and $x^3$ term since $P$ is odd. The $x$ term will clearly be $(p - 1)! x$; the $x^3$ term will be $-(p - 1)! x^3 \left(\frac{1}1 + \frac{1}2 + \cdots + \frac{1}{p - 1}\right)$. Then mod $p^4$, we have $\displaystyle \binom{p^2}{p} = p - p^3 \left(\frac{1}1 + \frac{1}2 + \cdots+ \frac{1}{p - 1}\right)$ (taking division modulo $p^4$ as well).
Note that when reducing $\mod p$, we have $\frac{1}1 + \frac{1}2 + \cdots + \frac{1}{p - 1} = 1 + 2 + ... + (p - 1)$, since every number from $1$ to $p - 1$ is a unit. And this sum is equal to $\frac{p (p - 1)}{2} \equiv 0 \pmod p$, since $p > 2$. Thus, we see that $\frac{1}1 + \frac{1}2 + \cdots + \frac{1}{p - 1}$ will be divisible by $p$ when the division is done $\mod p^4$ as well.
Thus, we have that for all $p>2$ prime, $\displaystyle \binom{p^2}{p} \equiv p \pmod {p^4}$.
I don't have anything going the other direction yet.
