show that x^4 == 2( mod p) has a solution for p==1(mod 4) iff(adsbygoogle = window.adsbygoogle || []).push({});

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 ?

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# 2 biquadratic residue mod p

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