Prime testing:-$f(p)={1\over p(p-2)}{p+1\choose p-3}$

Question

Another experiment to find some sort of function for test prime numbers

Conjecture:

Given that $$f(p)={1\over p(p-2)}{p+1\choose p-3}$$ Where p are prime numbers, $p\ge5$ then $f(p)$ will never have a remainder

Examples

$f(5)=1$

$f(7)=2$

$f(11)=5$

$f(13)=7$

How can one prove or disprove this conjecture?

Answer 1

The formula simplifies to $f(p)=(p+1)(p-1)/24$. This is an integer for any odd number which is not a multiple of $3$. (Since $p$ is odd, one of $(p-1)$ and $(p+1)$ is a multiple of $4$ and the other is even, and since $p$ is not a multiple of $3$, either $(p-1)$ or $(p+1)$ must be.)

So it's certainly true for primes greater than $3$, but it's useless for primality testing.