for a prime p of the form 4k1 For a prime p of the form 4K

For a prime p of the form 4K + 1, prove 1^1 middot 3^2 middot 5^2 (p - 2)^2 equivalence - 1 (mod p). Today is Thursday, what is the day of the week after 1111^1111 days?

Solution

By Wilson\'s Theorem for any prime p

(p-1)!=-1 mod p

(p-1)!=1*2*...*(p-2)*(p-1)=(1*(p-1))*...(k*(p-k))...=(1*(p-1))*(3*(p-3))...((p-2)*2)=(1*(-1))*(3*(-3))...*((p-2)*-(p-2))=(-1)^{(p-1)/2}1^23^2....(p-2)^2=-1

p=4k+1

So, (p-1)/2=2k

So,

Hence proved