# 2 biquadratic residue mod p

show that x^4 == 2( mod p) has a solution for p==1(mod 4) iff
p is of the form A^2+64B^2, where A,B are integers

I let x^2=M
then the conguence is reduced to M^2==2( mod p)
but any # squared == 0 or 1 ( mod4 ) so p must be == 1(mod4)...
but i'm not sure what to do now..
any hints/ or ideas ?