True or False(2)

Is it true or false that for any odd prime p p the following exists: 1 2 × 3 2 × 5 2 × × ( p 2 ) 2 ( 1 ) p + 1 2 m o d p 1^2\times3^2\times5^2\times\dots\times(p-2)^2\equiv(-1)^{\frac{p+1}{2}} \mod p

True False

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1 solution

Arousse Fares
Aug 11, 2018

We first put some order in this messy equation and exploit the fact that p + n n m o d p p+n \equiv n \mod p :

A ( p ) = 1 2 × 3 2 × . . . × ( p 2 ) 2 = 1 × ( p 2 ) × 3 × ( p 4 ) × . . . × ( p ( p 1 ) ) 1 × ( 2 ) × 3 × ( 4 ) × . . . × ( p 2 ) × ( ( p 1 ) ) m o d p A(p) = 1^{2} \times 3^{2} \times ... \times (p-2)^{2} = 1 \times (p-2) \times 3 \times (p-4) \times ... \times (p-(p-1)) \equiv 1 \times (-2) \times 3 \times (-4) \times ... \times (p-2) \times (-(p-1)) \mod p

Which means that if there is an odd number of even numbers then A ( p ) = ( p 1 ) ! A(p) = - (p-1)! and if there is an even number of even numbers then A ( p ) = ( p 1 ) ! A(p) = (p-1)!

According to Wilson's theorem : ( n 1 ) ! 1 m o d n (n-1)! \equiv -1 \mod n if and only if n n is prime.

And what a coincidence because we do have a prime number !!

The rest is easy. There are two cases : either you have an even number of even numbers or an odd number of even numbers. You just have to study the two scenarios with m o d 4 \mod 4 ...

Thank you for the problem ! It taught me Wilson's theorem which is quite a powerful tool.

Thank you, nice solution.

Hana Wehbi - 2 years, 10 months ago

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