# 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.

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## 1 Answer

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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I understood everything except "so 2 is a biquadratic residue iif b′ is even" how does $2^{(p-1)/4}=1$ imply that 2 is a biquadratic residue? Thanks! – quanta Mar 7 '11 at 23:31
2 is a biquadratic residue if and only if there is an $y$ with $y^4\equiv 2\pmod{p}$ if and only if $2^{(p-1)/4} \equiv (y^4)^{(p-1)/4} \equiv 1 \pmod{p}$ the last iif can be seen for example using a primitive root mod $p$. – Esteban Crespi Mar 8 '11 at 7:44