blah blah blah...
the problem reduces to computing ∑ z = 0 p − 1 ( c z − b z 2 ) ( p − 1 ) / 2 {\displaystyle \sum _{z=0}^{p-1}(cz-bz^{2})^{(p-1)/2}}
This is equal to: ∑ z = 0 p − 1 ∑ m = 0 ( p − 1 ) / 2 ( c z ) m ( − b z 2 ) ( p − 1 ) / 2 − m C m ( p − 1 ) / 2 {\displaystyle \sum _{z=0}^{p-1}\sum _{m=0}^{(p-1)/2}(cz)^{m}(-bz^{2})^{(p-1)/2-m}C_{m}^{(p-1)/2}} where C m n {\displaystyle C_{m}^{n}} is the combinations formula. Rearranging the equation gives:
∑ m = 0 ( p − 1 ) / 2 c m ( − b ) ( p − 1 ) / 2 − m C m ( p − 1 ) / 2 ∑ z = 0 p − 1 z ( p − 1 ) − 2 m {\displaystyle \sum _{m=0}^{(p-1)/2}c^{m}(-b)^{(p-1)/2-m}C_{m}^{(p-1)/2}\sum _{z=0}^{p-1}z^{(p-1)-2m}}
The only nonzero term is given when m=0 so we get ( p − 1 ) ( − b ) ( p − 1 ) / 2 {\displaystyle (p-1)(-b)^{(p-1)/2}} which modulo p gives − ( − b ) ( p − 1 ) / 2 = − ( − b / p ) {\displaystyle -(-b)^{(p-1)/2}=-(-b/p)}
where 
is the combinations formula. Rearranging the equation gives:
which modulo p gives 
