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, RRL Apr 4 at 16:20

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  • $\begingroup$ Use $73-a\equiv-a\pmod{73}$. $\endgroup$ – Lord Shark the Unknown Mar 31 at 6:45
  • $\begingroup$ $\binom{72}{k}\equiv(-1)^k\pmod{73}$ $\endgroup$ – robjohn Mar 31 at 9:26
  • $\begingroup$ Special case of this. $\endgroup$ – 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 $$73 | \left(\frac{(72!)}{(36!)^2}-1\right) \cdot (36!)^2$$

Since 73 is prime....


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