Prove that $\dfrac{72!}{(36!)^2}-1$ is divisible by $73$

Question

Prove that $\dfrac{72!}{(36!)^2}-1$ is divisible by $73$

My multiple attempts are as follows:-

Attempt $1$:

If we can prove $\dfrac{72!}{(36!)^2}$ as $73\lambda+1$, then we will be done

$$\dfrac{72!}{(36!)^2}={36\choose 0}^2+{36\choose 1}^2+{36\choose 2}^2+\cdots\cdots+{36\choose 36}^2$$

I didn't get anything from here, so I tried another method

Attempt $2$:

$$72!=2^{36}(36!)\prod_{k=1}^{36}(2k-1)$$

$$\dfrac{72!}{(36!)^2}=\dfrac{2^{36}\prod_{k=1}^{36}(2k-1)}{36!}$$ $$\dfrac{72!}{(36!)^2}=\dfrac{2^{36}\prod_{k=1}^{36}(2k-1)}{2^{18}(18!)\prod_{k=1}^{18}(2k-1)}$$

$$\dfrac{72!}{(36!)^2}=\dfrac{2^{36}\prod_{k=1}^{36}(2k-1)}{2^{27}(9!)\prod_{k=1}^{9}(2k-1)\prod_{k=1}^{18}(2k-1)}$$

$$\dfrac{72!}{(36!)^2}=\dfrac{2^{9}\prod_{k=1}^{36}(2k-1)}{9\cdot2^4(4!)\prod_{k=1}^{4}(2k-1)\prod_{k=1}^{9}(2k-1)\prod_{k=1}^{18}(2k-1)}$$

$$\dfrac{72!}{(36!)^2}=\dfrac{4}{27}\cdot\dfrac{\prod_{k=1}^{36}(2k-1)}{\prod_{k=1}^{4}(2k-1)\prod_{k=1}^{9}(2k-1)\prod_{k=1}^{18}(2k-1)}$$

$$\dfrac{72!}{(36!)^2}=\dfrac{4}{27}\cdot\dfrac{37\cdot39\cdot41\cdots\cdots71}{(1\cdot3\cdot5\cdot7)(1\cdot3\cdot5\cdot7\cdots\cdots17)}$$

Now it will take lot of time to cancel out all these terms, any other way of doing this problem.

Answer 1

Let us work in the field $F=\Bbb F_{73}$, with characteristic $73$, a prime number. Then Wilson's theorem shows $72!=-1$ in $F$, but we go an other way, and calculate in $F$: $$ \begin{aligned} (36!)^2 &=36!\;36!\\ &=36!\cdot(-36)(-35)\dots (-2)(-1)\\ &=36!\cdot(73-36)(73-35)\dots (73-2)(73-1)\\ &=72!\ . \end{aligned} $$ So we have to show $72!/72!-1=1-1$ is zero in $F$.

$\square$

Answer 2

$\displaystyle \binom{72}{36}=$ coefficient of $x^{36}$ in $\displaystyle (1-x)^{72} = \frac{(1-x)^{73}}{1-x}$

$$=(1-x)^{73}(1-x)^{-1}$$

$$=\bigg[\binom{73}{0}-\binom{73}{1}x+\binom{73}{2}x^2+\cdots \bigg](1+x+x^2+\cdots)$$

$$=\binom{73}{0}-\binom{73}{1}+\binom{73}{2}+\cdots +\binom{73}{36}$$

which is divisible by $73$

due to the fact that $\displaystyle \binom{p}{r}$ is divisible by $p$

$p$ is prime number and $r\in \{1,2,3,4,\cdots,p-1\}$

Answer 3

$$\frac{72!}{(36!)^2}-1=\frac{36!37.38...72}{(36)^2}-1$$ $$=\frac{37.38.39.....72}{36!}-1$$ $$=\frac{(73-36)(73-35)....(73-1)}{36!}-1$$ $$=\frac{73m+36!}{36!}-1$$ $$=\frac{73m}{36!}+1-1$$ $$=\frac{73m}{36!}=73n \,where\, n\in N$$ Where m will be multiple of $(36!)$ as $\frac{72!}{(36!)^2}={72 \choose 36} \in N$