# Prove that $\frac{(72!)}{(36!)^2}-1$ is divisible by 73 [closed]

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

My approach is as follow $$73n=\frac{(72!)}{(36!)^2}-1$$ I tried remainder theorem but could not prove it.

## closed as off-topic by user21820, José Carlos Santos, Adrian Keister, K.Power, RRLApr 4 at 16:20

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• Use $73-a\equiv-a\pmod{73}$. – Lord Shark the Unknown Mar 31 at 6:45
• $\binom{72}{k}\equiv(-1)^k\pmod{73}$ – robjohn Mar 31 at 9:26
• Special case of this. – Bill Dubuque Apr 1 at 3:04

Because $$\frac{72!}{(36!)^2}=\frac{72\cdot71\cdot...\cdot37}{1\cdot2\cdot...\cdot36}\equiv\frac{-1\cdot(-2)\cdot...\cdot(-36)}{1\cdot2\cdot...\cdot36}=(-1)^{36}=1.$$
Hint: Prove that $$72! \equiv (36!)^2 \pmod{73}$$
Hint 2: Deduce from above that $$73 | \left(\frac{(72!)}{(36!)^2}-1\right) \cdot (36!)^2$$