Let p be an odd prime Let T p12 that T Solutionx1p12 Note Fo

Let p be an odd prime. Let T (p-1)/2)!. that T

Solution

x=1*....*(p-1)/2

Note,

For every 1<=k<=(p-1)/2 there is a corresponding p-1>=p-k>=(p+1)/2

SO that

p-k=-k mod p

x^2=1*1*....*(p-1)/2*(p-1)/2=1*(-(p-1))....*k*(-(p-k))*...*(p-1)/2*(-(p+1)/2

                  =(-1)^{(p-1)/2}(p-1)! modulo p

p=1 mod 4

p=4k+1

(p-1)/2=2k

SO,

(-1)^{(p-1)/2}=1

So

x^2=(p-1)! modulo p

By Wilson\'s Theorem

(p-1)!=-1 modulo p

Hence

x^2=-1 modulo p