It seems that Gauss states the following theorem in his first paper of biquadratic residues(Werke vol. II pp. 67-92). I cannot read Latin, but I have a Japanese translation of the paper. However, it is difficult to decipher the paper even in Japanese. Is the theorem right? If yes, how do you prove it?

Theorem Let $p$ be a prime of the form $p = 4n + 1$. Let $g$ be a primitive root mod $p$. Let $f \equiv g^{(p-1)/4}$ (mod $p$). Then $f^2 \equiv -1$ (mod $p$). It is well known that $p = x^2 + y^2$ has an integer solution $(a, b)$. Suppose $a$ is odd and $b$ is even. $a$ is uniquely determined by the condition $a \equiv 1$ (mod $4$). $b$ is uniquely determined by the condition $b \equiv af$ (mod $p$). Suppose $2 \equiv g^\lambda$ (mod $p$). Then $\lambda \equiv b/2$ (mod $4$).

Remark Here's a quote from Weil's famous paper "Number of solutions of equations in finite fields". The highlighting is mine.

He draws attention himself to the elegance of his method, as well as to its wide scope; it is only much later, however, viz. in his first memoir on biquadratic residues [lb], that he gave in print another application of the same method ; there he treats the next higher case, finds the number of solutions of any congruence $ax^4 — by^4 \equiv 1$ (mod $p$), for a prime of the form $p = 4n + l$, and derives from this the biquadratic character of $2$ mod $p$, this being the ostensible purpose of the whole highly ingenious and intricate investigation.

Gauss's Werke vol. II p.89

Summa harum investigationum ita enunciari potest: Numerus 2 pertinet ad complexum $A, B, C$ vel $D$, prout numerus $b/2$ est formae $4m, 4m + 1, 4m + 2$ vel $4m +3$.

[Translation] The result of the investigation can be described as follows: Number $2$ belongs to the set $A, B, C$ or $D$ according $b/2$ is of the form $4m, 4m + 1, 4m +2$ or $4m + 3$.


One way to think about this question is to consider the extension $\mathbb Q(i,2^{1/4})$, which is an abelian extension of $\mathbb Q(i)$, and then a class field of some conductor. What is the conductor?

Well, it is unram. ouside $1+i$, so the conductor is some power of $(1+i)$.

The ray class field of conductor $(1+i)^3$ is just $\mathbb Q(i)$ itself. The ray class field of conductor $(i+i)^4$ is $\mathbb Q(i,\sqrt{2}) = \mathbb Q(\zeta_8)$. The ray class field of conductor $(i+i)^5$ is $\mathbb Q(\zeta_8,\sqrt{1+i})$. The ray class field of conductor $(1+i)^6$ is $\mathbb Q(\zeta_{16}, 2^{1/4}).$ (You can check these using e.g. the conductor-discriminant formula.)

So $\mathbb Q(i,2^{1/4})$ has conductor $(1+i)^6$.

Now if $p = a^2 + b^2$ ($p$ a prime that is $1 \bmod 4$), then we can arrange that $a$ is odd and $b$ even, and that $a - b \equiv 1 \bmod 4$.

If we write $\pi = a + b i $, then these conditions on $a$ and $b$ are equivalent to $\pi \equiv 1 \bmod (1+i)^3$.

In order to have $\pi \equiv 1 \bmod (1+i)^4$, we need $a \equiv 1 \bmod 4$ and $b \equiv 0 \bmod 4$.

In order to have $\pi \equiv 1 \bmod (1+i)^5$, we need $a \equiv 1 \bmod 8$ and $b \equiv 0 \bmod 4$.

In order to have $\pi \equiv 1 \bmod (1+i)^6 $, we need $a \equiv 1 \bmod 8$ and $b \equiv 0 \bmod 8$.

The fact that $\mathbb Q(i,2^{1/4})$ has conductor $(1+i)^6$ then tells us, for example, that the biquadratic character of $2$ mod $\pi$ (or mod $p$) will depend on $a$ and $b$ mod $8$, and not just on (say) $a \bmod 8$ and $b \bmod 4$.

This doesn't give the precise answer; for that we would have to actually compute some Artin symbols. (And this is how one would deduce biquadratic reciprocity from general class field theory.) But it may help to explain why the result is true.

  • $\begingroup$ Thanks. This is interesting. I guess I need some time to assimilate this, though. $\endgroup$ – Makoto Kato Jan 4 '14 at 21:52

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