\documentclass[12pt]{article}
\include{macros}
\title{Lecture 21: Binary Quadratic Forms I:\\
Sums of Two Squares}
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\maketitle
Today we study the question of which integers are the sum of two squares.

\section{Sums of Two Squares}
During the next four lectures, we will study binary quadratic forms.
A simple example of a binary quadratic form that will occupy us today is 
$$
  x^2 + y^2.
$$
A typical question that one asks about a quadratic form is which
integers does it represent.  ``Are there integers~$x$ and~$y$ 
so that $x^2 + y^2 = 389$?  So that $x^2 + y^2 = 2001$?''


\subsection{Which Numbers are the Sum of Two Squares?}
The main goal of today's lecture is to prove the following theorem.
\begin{theorem}\label{thm:sumsquare}
A number~$n$ is a sum of two squares if and only if all prime factors
of~$n$ of the form $4m+3$ have even exponent in the prime factorization
of~$n$.
\end{theorem}
Before tackling a proof, we consider a few examples.
\begin{example}\mbox{}\vspace{-3ex}\\
\begin{itemize}
\item $5 = 1^2 + 2^2$.
\item $7$ is not a sum of two squares.
\item $2001$ is divisible by $3$ because $2+1$ is, but not by $9$ since $2+1$ is not, so $2001$ is {\em not} a sum of two squares.
\item $2\cdot 3^4\cdot 5\cdot 7^2\cdot 13$ is a sum of two squares.
\item $389$ is a sum of two squares, since $389\con 1\pmod{4}$ and $389$ is prime.
\item $21=3\cdot 7$ is {\em not} a sum of two squares even though $21\con 1\pmod{4}$.
\end{itemize}
\end{example}

In preparation for the proof of Theorem~\ref{thm:sumsquare}, we recall a
result that emerged when we analyzed how  partial convergents
of a continued fraction converge.
\begin{lemma}\label{lem:approx}
If $x\in\R$ and $n\in\N$, then there is a fraction $\ds\frac{a}{b}$
in lowest terms such that $0<b\leq n$ and
$$\left| x - \frac{a}{b} \right| \leq \frac{1}{b(n+1)}.$$
\end{lemma}
\begin{proof}
Let $[a_0,a_1,\ldots]$ be the continued fraction expansion of~$x$.
As we saw in the proof of Theorem~2.3 in Lecture~18, for each~$m$
$$
 \left| x - \frac{p_m}{q_m}\right|
  < \frac{1}{q_m \cdot q_{m+1}}.
$$
Since $q_{m+1}$ is always at least~$1$ bigger than $q_m$ and $q_0=1$, 
either there exists an~$m$ such that $q_m\leq n < q_{m+1}$, or the
continued fraction expansion of~$x$ is finite and $n$ is larger
than the denominator of the rational number~$x$.  In the first
case,
$$
  \left| x - \frac{p_m}{q_m}\right|
   < \frac{1}{q_m \cdot q_{m+1}}
      \leq \frac{1}{q_m \cdot (n+1)},$$
so $\ds\frac{a}{b} = \frac{p_m}{q_m}$ satisfies the conclusion of
the lemma.   In the second case, just let $\ds\frac{a}{b} = x$.

\end{proof}

\begin{definition}
A representation $n=x^2 + y^2$ is {\em primitive} if
$\gcd(x,y)=1$.%; otherwise, it is {\em imprimitive}.
\end{definition}

\begin{lemma}\label{lem:primitive}
If~$n$ is divisible by a prime~$p$ of the form $4m+3$, then~$n$
has no primitive representations.
\end{lemma}
\begin{proof}
If~$n$ has a primitive representation, $n=x^2 + y^2$, then
$$
   p \mid x^2 + y^2\quad \text{ and }\quad \gcd(x,y)=1,
$$
so $p\nmid x$ and $p\nmid y$.  Thus
$x^2 + y^2 \con 0\pmod{p}$
so, since $\Z/p\Z$ is a field we can divide by $y^2$ and see that
$$
  (x/y)^2 \con -1\pmod{p}.
$$
Thus the quadratic residue symbol $\kr{-1}{p}$ equals $+1$.
However,
$$
   \kr{-1}{p} = (-1)^{\frac{p-1}{2}} = (-1)^\frac{4m+3-1}{2} = (-1)^{2m+1} = -1.
$$
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:sumsquare}]
\noindent$\left(\Longrightarrow\right)$
Suppose that~$p$ is of the form $4m+3$, that $p^r\mid\mid n$ (exactly
divides) with~$r$ odd, and that $n=x^2+y^2$.  Letting $d=\gcd(x,y)$,
we have
$$
   x = dx', \quad y = dy', \quad n = d^2 n'
$$
with $\gcd(x',y')=1$ and
$$
   (x')^2 + (y')^2 = n'.
$$

Because~$r$ is odd, $p\mid n'$, so Lemma~\ref{lem:primitive}
implies that $\gcd(x',y')>1$, a contradiction.
\vspace{0.3ex}

\noindent$\left(\Longleftarrow\right)$
Write $n=n_1^2 n_2$ where $n_2$ has no prime factors of the 
form $4m+3$.  It suffices to show that~$n_2$ is a sum of two
squares.  Also note that
$$
  (x_1^2 + y_1^2)(x_2^2+y_2^2) = (x_1x_2+y_1y_2)^2 + (x_1y_2-x_2y_1)^2,
$$
so a product of two numbers that are sums of two squares is also
a sum of two squares.\footnote{This algebraic identity is secretely
the assertion that the norm map $N:\Q(i)^* \ra \Q^*$ sending
$x+iy$ to $(x+iy)(x-iy)=x^2+y^2$ is a homomorphism.}
Also, the prime~$2$ is a sum of two squares.
It thus suffices to show that if~$p$ is a prime of the form
$4m+1$, then~$p$ is a sum of two squares.

Since 
$$
  (-1)^{\frac{p-1}{2}} = (-1)^{\frac{4m+1-1}{2}} = +1,
$$
$-1$ is a square modulo~$p$; i.e., there exists~$r$ such
that $r^2\con -1\pmod{p}$.  Taking $n=\lfloor \sqrt{p}\rfloor$
in Lemma~\ref{lem:approx} we see that there are integers $a, b$
such that
$0<b<\sqrt{p}$ and 
$$
 \left| -\frac{r}{p} - \frac{a}{b}\right| \leq\frac{1}{b(n+1)} < \frac{1}{b\sqrt{p}}.
$$
If we write
$$
  c = rb + pa
$$
then
$$
  |c| < \frac{pb}{b\sqrt{p}} = \frac{p}{\sqrt{p}} = \sqrt{p}
$$
and
$$
   0 < b^2 + c^2 < 2p.
$$
But $c \con rb\pmod{p}$, so 
$$
  b^2 + c^2 \con b^2 + r^2 b^2 \con b^2(1+r^2) \con 0\pmod{p}.
$$
Thus $b^2 + c^2 = p$.
\end{proof}

\subsection{Computing $x$ and $y$}
Suppose~$p$ is a prime of the form $4m+1$.  There is a construction of
Legendre of~$x$ and~$y$ that is explained on pages 120--121 of
Davenport.  I'm unconvinced that it is any more efficient than the
following naive algorithm: compute $\sqrt{p-x^2}$ for $x=1,2,\ldots$
until it's an integer.  This takes at most $\sqrt{p}$ steps.  Here's
a simple PARI program which implements this algorithm.

\begin{verbatim}
{sumoftwosquares(n) =
   local(y);
   for(x=1,floor(sqrt(n)),
      y=sqrt(n-x^2); 
      if(y-floor(y)==0, return([x,floor(y)]))
   );
   error(n," is not a sum of two squares.")
}
\end{verbatim}

\section{Sums of More Squares}
Every natural number is a sum of {\bf four} squares.  See pages
124--126 of Davenport for a proof.  

A natural number is a sum of {\bf three} squares if and only if it is not a
power of~$4$ times a number that is congruent to~$7$ modulo~$8$.  For
example, $7$ is not a sum of three squares.  This is more difficult to
prove.


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