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THEORY OF NUMBERS

SUMS OF SQUARES

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We wish to know which numbers are representable as sums of
squares. For two squares

$$(x_1^2+y_1^2)(x_2^2+y_2^2)=(x_1x_2+y_1y_2)^2+(x_1y_2-x_2y_1)^2$$

So the product of two numbers representable is also representable.
We then ask what primes are representable.

Now the square of any even number is congruent to 0 (mod 4) and
the square of any odd number is $\equiv 1\ (4)$. So the sum of two
squares is congruent to 0 1 or 2 mod 4, so any number $4m+3$ is
not representable.

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\item[Theorem (Fermat)]
Every prime $p\equiv1\ (4)$ is representable as the sum of two
squares.

\item[Proof 1]
$\exists x_0$ such that $x_0^2+1\equiv 0$ mod $p$. $(-1)$ is a
quadratic residue mod $p$ therefore $mp=x^2+y^2$ for some $m\
(x_0^2+1^2-mp)$.

Let $m$ be the least positive integer such that $mp=x^2+y^2$. Then
$1\leq m\leq p.$ R.T.P. $m=1$

Assume $m>1,\ \exists x_1,\ y_1$ such that $x\equiv x_1$ mod $m\
y\equiv y_1$ mod $m$.

$|x_1|\leq\frac{1}{2}m\ |y_1|\leq\frac{1}{2}m\\ x_1^2+y_1^2\equiv
x^2+y^2=0$ mod $m$ Therefore $lm=x_1^2+y_1^2$ where $l$ is an
integer $l=0\Rightarrow x_1^2=y_1^2=0\Rightarrow x\equiv0\
y\equiv0\Rightarrow m|x\ m|y\Rightarrow m|p$ therefore $1\leq l<m$

For
$lm\leq\left(\frac{1}{2}m\right)^2+\left(\frac{1}{2}m\right)^2=\frac{1}{2}m^2<m^2$

Now $lpm^2=(x^2+y^2)(x_1^2+y_1^2)=(xx_1+yy_1)^2+(xy_1-x_1y)^2$

Now $xx_1+yy_1\equiv0$ mod $m$ and $xy_1-x_1y\equiv 0$ mod $m$
therefore $lp=u^2+v^2,\ v,\ v$ integers. This is a contradiction
therefore $m=1$.

\item[Proof 2]
$\exists\lambda$ such that $\lambda^2+1\equiv0$ mod $p$ S.T.P.
$\exists (x,y)\neq(0,0)$ such that $y\equiv\lambda x$ mod $p\
x^2+y^2<2p$.

\item[Lemma]
Suppose $\lambda\not\equiv0$ mod $p$. Suppose $e,\ f$ are natural
numbers such that $ef>p$ then $\exists$ a non-trivial solution
$x,\ y$ of $y\equiv\lambda x$ mod $p$ satisfying $|x|\leq e-1\
|y|\leq f-1$.

\item[Proof]
Consider the set $S$ of $y-\lambda x$ as $x,\ y$ run through
$0,1,\ldots e-1;\ 0,1,\ldots f-y$. The number of elements in $s$
is $ef>p$ therefore $\exists\ y'-\lambda x'\equiv y''\lambda x''$
mod $p$ such that $(x'-x'')^2+(y'-y'')^2\neq0$ Put $y=y'-y'';\ m
x=x'-x'''$ then $x$ and $y$ are the required numbers.

Now apply the lemma with $e=f=[p^\frac{1}{2}]+1>p^\frac{1}{2}$
then $ef>p$ and $(e-1)^2+(f-1)^2=2[p^\frac{1}{2}]^2<2p$.

\item[Theorem]
A natural number $n$ is representable as the sum of two squares
$\Leftrightarrow$ every prime $q\equiv-1$ mod 4 which divides $n$
divides it to an even power.

\item[Proof]
S.C. obvious.

N.C. suppose $n=x^2+y^2$ and $q|n$ where $q=-1\ (4)$.

Suppose $q\not|x\\ x^2+y^2\equiv0$ mod $q$ and $\exists x_0$ such
that $xx_0\equiv1$ mod $q$. Therefore $(x_0y)^2\equiv-1$ mod $q$.

i.e. $-1$ is a quadratic residue mod $q$ which is false therefore
$q|x$ and $q|y$ therefore $q^2|n$ therefore
$\frac{n}{q^2}=x_1^2+y_1^2$.

If $q|\frac{n}{q^2}$ we repeat the argument. We can only do so a
finite number of times and so $q$ divides $n$ to an even power.

Considering sums of 4 squares we have the following identity

\begin{eqnarray*}
(x_1^2+x_2^2+x_3^2+x_4^2)(y_1^2+y_2^2+y_3^2+y_4^2)
&=&(x_1y_1+x_2y_2+x_3y_3+x_4y_4)^2\\
&+&(x_1y_2-x_2y_1+x_3y_4-x_4y_3)^2\\
&+&(x_1y_3-x_3y_1-x_2y_4+x_4y_2)^2\\
&+&(x_1y_4-x_4y_1-x_3y_2+x_2y_3)^2
\end{eqnarray*}

From which it follows that the product of two representable
numbers is representable.

Again if $x_i\equiv y_i$ mod $m,\ i=1,2,3,4$ and $\sum
x_i^2\equiv0$ then each of the four expressions on the right hand
sided is $\equiv0$ mod $m$.

\item[Theorem (Lagrange)]
Every natural number is representable as the sum of four squares.

\item[Proof]
S.T.P. for primes by the above identity.

$2=1^2+1^2+0^2+0^2\\ p\equiv1\ (4)\ p^2=x^2+y^2+0^2+0^2$

S.T.P. for $q\equiv -1\ (4)$

$\exists$ an integer $a$ such that $\left(\frac{a}{q}\right)\
\left(\frac{a+1}{q}\right)=-1$

Then since $q\equiv-1\ (4)$ we have
$\left(\frac{-a-1}{q}\right)=-\left(\frac{-1}{q}\right)=+1$.

$\exists x_1$ such that $x_1^2\equiv a\ (q)$ and $\exists x_2$
such that $x_2^2\equiv-a-1\ (q)$.

Now $x_1^2+x_2^2+1^2+0^2\equiv0$ mod $q$ so some non-zero multiple
of $q$ is representable.

Let $m$ be the least positive integer such that
$mq=x_1^2+x_2^2+x_3^2+x_4^2$. Then $1\leq m<q$. Suppose that $m>1$

We first prove that $m$ is odd. Suppose that $m$ is even. Then the
number of odd $x$'s is even, and suppose that they come first in
the representation.

$$\left(\frac{1}{2}m\right)q=\left(\frac{x_1+x_2}{2}\right)^2
+\left(\frac{x_1-x_2}{2}\right)^2+\left(\frac{x_3+x_4}{2}\right)^2
+\left(\frac{x_3-x_4}{2}\right)^2$$

All the terms on the right hand side are integers so we have a
contradiction, since $\left(\frac{1}{2}m\right)q$ is not
representable.

Thus $m$ is odd, and so for $i=1,2,3,4$ we choose $y_i$ such that
$x_i\equiv y_i\ (m)\ |y_i|<\frac{1}{2}m$ then

$lm=y_1^2+y_2^2+y_3^2+y_4^2$ where $1\leq l<m$

So
$lm^2q=(x_1^2+x_2^2+x_3^2+x_4^2)(y_1^2+y_2^2+y_3^2+y_4^2)=A^2+B^2+C^2+D^2$
by the above identity.

$A,\ B,\ C,\ D$ are all divisible by $m$ and so $lq$ is
representable. Thus we have a contradiction and so $m=1$.

For 3 squares the result is $a^q(8t+7)$ not representable, all
others are.

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