<s>[INST] Help complete a proof of Dirichlet on biquadratic character of 2? I am stuck proving the theorem that there exists $x$, $x^4 \equiv 2 \pmod p$ iff $p$ is of the form $A^2 + 64B^2$.
So far I have got this (and I am not sure if it's correct)

Let $p = a^2 + b^2$ be an odd prime,


*

*$\left(\frac{a}{p}\right) = \left(\frac{p}{a}\right) = \left(\frac{a^2 + b^2}{a}\right) = \left(\frac{b^2}{a}\right) = 1$


since $p \equiv 1 \pmod 4$


*

*$\left(\frac{a+b}{p}\right) = \left(\frac{(a+b)^2-2ab}{a+b}\right) = \left(\frac{2}{a+b}\right) = (-1)^{((a+b)^2-1)/8}$


using the Jacobi symbol and second supliment of quadratic reciprocity.


*

*$(a+b)^{(p-1)/2} = (2ab)^{(p-1)/4}$


since $(a+b)^2 \equiv 2ab \pmod p$

and the last step which I'm stuck on now is for $p = a^2 + b^2$ let $x^2 \equiv -1 \pmod p$ then $2^{(p-1)/4} = x^{ab/2}$. And I don't see how to prove the theorem with this result.
 [/INST] We can asume that 2 is a quadratic residue mod $p$ and so that $p \equiv 1 \pmod 8$ and this implies that if we pick $a$ odd and $b$ even then $b$ is a multiple of 4. We have to prove that $b$ is a multiple of 8. 
First observe that as $x^2 \equiv -1 \pmod{p}$ and $a^2 + b^2 = p$ we have 
    $$ \left(\frac{a+b}{p}\right) \equiv (-1)^{((a+b)^2-1)/8} \equiv x^{(p+2ab-1)/4} \pmod{p} $$
note that the exponent of $x$ is even because $b$ is multiple of 4, so we can chose the sign of the base as we wish.
In adition we also have 
    $$ -a^2 \equiv b^2 \pmod{p} $$
so $(xa)^2  \equiv b^2 \pmod{p}$ and 
   $$ax \equiv \pm b \pmod{p}$$
and picking the sign of $x$ we can asume that $ax \equiv b \pmod{p}$. So  $ab \equiv a^2 x$ and 
  $$ (ab)^{(p-1)/4} = a^{(p-1)/2} x^{(p-1)/4} \equiv x^{(p-1)/4} \pmod{p} $$
because $a$ is a quadratic residue mod $p$.  
The identity you obtained:
    $$\left(\frac{a+b}{p}\right) = (a+b)^{(p-1)/2} \equiv (2ab)^{(p-1)/4} \pmod p $$
now becomes
    $$x^{(p+2ab-1)/4} \equiv 2^{(p-1)/4} x^{(p-1)/4} \pmod p $$
and in consequence
    $$2^{(p-1)/4} \equiv x^{ab/2} \pmod p $$
As $b$ is multiple of 4 say $b = 4b'$, we have
    $$2^{(p-1)/4} \equiv (-1)^{ab'} \pmod p $$
so $2$ is a biquadratic residue iif $b'$ is even or what is the same thing iif $b$ is multiple of 8 and we are done. 
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