Is $72!/36! -1$ divisible by 73?

Question

Is $\frac{72!}{36!}-1$ divisible by the number 73?

I am not getting a clue in which direction should I go, though I did small amount of work by converting the above expression in the below given form

$$(1.3.5.7.9...69.71).2^{36} - 1$$

and

$$36!\binom{72}{36} -1$$

I am unable to proceed further.

Answer 1

Let $p=73$. $p$ is a prime number. Thus by Wilson's theorem $$72! = (p-1)!\equiv -1 \mod p$$

Notice that $\forall 1\le n\le 36$, $73-n \equiv -n \mod p$, thus

$$ \frac{72!}{36!}\equiv (37\cdot 38\cdots 72)\equiv (-1\cdot -2\cdots -36) \equiv (1\cdot 2\cdots 36)(-1)^{36} \equiv (1\cdot 2\cdots 36) \equiv 36! \mod p$$

Suppose for the sake of contradiction that $ \frac{72!}{36!} \equiv 1 \mod p$, then $ 36! \equiv 1 \mod p$ and

$$ 72! \equiv (36!)^2 \equiv 1 \mod p$$

which is not true.

So we can conclude that $\frac{72!}{36!} \mod p$ is not $1$.

Answer 2

$$\dfrac{(2m)!}{m!}=(2m)(2m-2)\cdots4\cdot2$$ $$\equiv(-1)(-3)\cdots(2m+1-4)(2m+1-2)\pmod{2m+1}$$ $$\equiv(-1)^m1\cdot3\cdots(2m-3)(2m-1)$$

Now if $2m+1$ is prime, $$(2m)!\equiv-1\pmod{2m+1}$$

$$-1\equiv(-1)^m\{(2m)(2m-2)\cdots4\cdot2\}^2$$

If $m=2n,$ $$\{(4n)(4n-2)\cdots4\cdot2\}^2\equiv-1\pmod{4n+1}$$

$$\implies(4n)(4n-2)\cdots4\cdot2\not\equiv\pm1\pmod{4n+1}$$