\documentclass[10pt]{amsart}

\title{The Biquadratic Character of 2}
\author{Jeff Garrett}
\date{February 20, 2001}

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\begin{document}
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  \maketitle
  \begin{thm}
    The primes $p \equiv \pmod 4$ for which $2$ is a biquadratic residue modulo $p$ are exactly
    those of the form $A^2 + 64 B^2$.
  \end{thm}
  \begin{proof}
    Suppose that $p \equiv 1 \pmod 4$.  Write $p = a^2 + b^2$ with $a$ odd.  The
    idea is to compute $(a+b/p)$ by two different methods.  Let $f$ be such that
    $b \equiv af \pmod p$.  Then $p = a^2 + b^2 = a^2 (1 + f^2) \equiv 0 \pmod p$
    and so $f^2 \equiv -1 \pmod p$.

    First, note that $(a+b)^2 + (a-b)^2 = 2p$ so that $1 = (2p/a+b) = (2/a+b)(p/a+b) 
    = (-1)^{((a+b)^2 - 1)/8} (p/a+b)$.  Thus $(a+b/p) = (-1)^{(a+b-1)(p-1)/4} (p/a+b) 
    = (-1)^{((a+b)^2 - 1)/8} \equiv f^{((a+b)^2 - 1)/4} \pmod p$.

    Second, since $(a+b)^2 \equiv 2ab \pmod p$, $(a+b/p) \equiv (a+b)^{(p-1)/2}
    \equiv (2ab)^{(p-1)/4} \equiv (2f)^{(p-1)/4} \pmod p$ since $(a/p) =
    (-1)^{(p-1)(a-1)/4} (p/a) = 1$.

    Equating the two expressions, we have $2^{(p-1)/4} \equiv f^{ab/2} \pmod p$.
    $2$ is a biquadratic residue iff this is $1$, but since $f$ is of order $4$,
    this happens iff $8$ divides $b$, i.e. $p$ is of the form required. 
  \end{proof}
\end{document}