x^2+y^2=1 ( mod p ) p is odd prime(4k+1) ,get all solutions ,
e^{2\pi i}=1
since cos^2\frac{{2\pi}}{n} +sin^2\frac{{2\pi}}{n}=1
i^2=-1 ( mod p )
I want to use this solve x^n=1 ( mod p ) for example,
x^5=1 (mod 41 ) x^2+y^2=1 ( mod 41 ) has 6 solutions (not include trivialsolution )
9^2=-1( mod 41) and  i=9 ( mod 41 )
so (x+iy)^5=1 (mod 41) (x+9y)^5=1 (mod 41 )
or (y+9x)^5=1 (mod 41 ) is this effective ?