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