# The biquadratic character of $2$ mod $p$ for a prime of the form $p=4n+1$ [closed]

It seems that Gauss states the following theorem in his first paper on 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 I asked the same question in Math StackExchange. Since nobody answered it so far and it seems that the question is highly non-trivial, I post this question here. Since I was suspended recently there, I'd appreciate if somebody would kindly add the link of this question to my question there.

This is a related question in MSE: https://math.stackexchange.com/questions/470976/the-number-of-solutions-of-ax4-by4-equiv-1-mod-p-for-a-prime-of-the-f

• I can't say that I'm a fan of using MO to circumvent MSE suspensions. – Cam McLeman Aug 20 '13 at 19:51
• @CamMcLeman What exactly do you mean by "to circumvent MSE suspension"? Does it mean using MO instead of MSE? What's wrong with it as long as questions are appropriate here? – Makoto Kato Aug 20 '13 at 21:29
• Some people downvoted for this and my other questions which were posted today. If they did it just because I was suspended in MSE, that would be outrageous. MO is NOT MSE. – Makoto Kato Aug 21 '13 at 1:37
• And maybe now people will not even have to read those threads to get an idea of the cause for suspension, seeing as there are 10 of them, most of them quite recent (within a few days). – Tobias Kildetoft Aug 21 '13 at 7:20
• You generally cause a lot of noise by creating multiple threads on subjects that have already been discussed previously at length without even bothering to acknowledge that previous discussion or apparently reading it. – Tobias Kildetoft Aug 22 '13 at 9:57

From quartic reciprocity we obtain $(2/p)_4 = i^{ab/2}$, or in other words, $2^{(p-1)/4} \equiv f^{ab/2} \pmod p$. (There is an elementary proof of this fact due to Dirichlet: see Exercise 4.24 in Cox, "Primes of the Form $x^2+ny^2$.) So if $2 \equiv g^\lambda \pmod p$, then $f^{ab/2} \equiv g^{\lambda(p-1)/4} \equiv f^\lambda \pmod p$, and consequently $\lambda - ab/2 \equiv 0 \pmod 4$.