Theorem 1.1 A number
is a sum of two squares if and only if all prime factors of
of the form
have even exponent in the prime factorization of
.
Before tackling a proof, we consider a few examples.
![$ n$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img7.png)
![$ n$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img7.png)
![$ 4m+3$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img8.png)
![$ n$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img7.png)
Example 1.2
.
is not a sum of two squares.
is divisible by
because
is, but not by
since
is not, so
is not a sum of two squares.
is a sum of two squares.
is a sum of two squares, since
and
is prime.
is not a sum of two squares even though
.
In preparation for the proof of Theorem 1.1, we recall a result that emerged when we analyzed how partial convergents of a continued fraction converge.
Proof. Let
be the continued fraction expansion of
. As we saw in the proof of Theorem 2.3 in Lecture 18, for each
Since
is always at least
bigger than
and
, either there exists an
such that
, or the continued fraction expansion of
is finite and
is larger than the denominator of the rational number
. In the first case,
so
satisfies the conclusion of the lemma. In the second case, just let
.
![$ [a_0,a_1,/ldots]$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img26.png)
![$ x$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img1.png)
![$ m$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img27.png)
![$/displaystyle /left/vert x - /frac{p_m}{q_m}/right/vert
< /frac{1}{q_m /cdot q_{m+1}}.
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img28.png)
![$ q_{m+1}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img29.png)
![$ 1$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img30.png)
![$ q_m$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img31.png)
![$ q_0=1$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img32.png)
![$ m$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img27.png)
![$ q_m/leq n < q_{m+1}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img33.png)
![$ x$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img1.png)
![$ n$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img7.png)
![$ x$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img1.png)
![$/displaystyle /left/vert x - /frac{p_m}{q_m}/right/vert
< /frac{1}{q_m /cdot q_{m+1}}
/leq /frac{1}{q_m /cdot (n+1)},$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img34.png)
![$ /displaystyle /frac{a}{b} = /frac{p_m}{q_m}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img35.png)
![$ /displaystyle /frac{a}{b} = x$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img36.png)
Definition 1.4 A representation
is primitive if
.
![$ n=x^2 + y^2$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img37.png)
![$ /gcd(x,y)=1$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img38.png)
Proof. If
has a primitive representation,
, then
and
so
and
. Thus
so, since
is a field we can divide by
and see that
Thus the quadratic residue symbol
equals
. However,
![$ n$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img7.png)
![$ n=x^2 + y^2$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img37.png)
![$/displaystyle p /mid x^2 + y^2$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img40.png)
![$/displaystyle /quad /gcd(x,y)=1,
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img41.png)
![$ p/nmid x$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img42.png)
![$ p/nmid y$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img43.png)
![$ x^2 + y^2 /equiv 0/pmod{p}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img44.png)
![$ /mathbb{Z}/p/mathbb{Z}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img45.png)
![$ y^2$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img46.png)
![$/displaystyle (x/y)^2 /equiv -1/pmod{p}.
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img47.png)
![$ /left(/frac{-1}{p}/right)$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img48.png)
![$ +1$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img49.png)
![$/displaystyle /left(/frac{-1}{p}/right) = (-1)^{/frac{p-1}{2}} = (-1)^/frac{4m+3-1}{2} = (-1)^{2m+1} = -1.
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img50.png)
![$ /qedsymbol$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img25.png)
Proof. [Proof of Theorem 1.1]
Suppose that
is of the form
, that
(exactly divides) with
odd, and that
. Letting
, we have
with
and
so a product of two numbers that are sums of two squares is also a sum of two squares.1Also, the prime
is a sum of two squares. It thus suffices to show that if
is a prime of the form
, then
is a sum of two squares.
is a square modulo
; i.e., there exists
such that
. Taking
in Lemma 1.3 we see that there are integers
such that
and
If we write
then
and
But
, so
Thus
.
![$ /left(/Longrightarrow/right)$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img51.png)
![$ p$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img39.png)
![$ 4m+3$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img8.png)
![$ p^r/mid/mid n$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img52.png)
![$ r$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img53.png)
![$ n=x^2 + y^2$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img37.png)
![$ d=/gcd(x,y)$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img54.png)
![$/displaystyle x = dx', /quad y = dy', /quad n = d^2 n'
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img55.png)
![$ /gcd(x',y')=1$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img56.png)
![$/displaystyle (x')^2 + (y')^2 = n'.
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img57.png)
Because is odd,
, so Lemma 1.5 implies that
, a contradiction.
Write
where
has no prime factors of the form
. It suffices to show that
is a sum of two squares. Also note that
![$/displaystyle (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,
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img63.png)
![$ 2$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img67.png)
![$ p$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img39.png)
![$ 4m+1$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img68.png)
![$ p$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img39.png)
Since
![$/displaystyle (-1)^{/frac{p-1}{2}} = (-1)^{/frac{4m+1-1}{2}} = +1,
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img69.png)
![$ -1$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img70.png)
![$ p$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img39.png)
![$ r$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img53.png)
![$ r^2/equiv -1/pmod{p}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img71.png)
![$ n=/lfloor /sqrt{p}/rfloor$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img72.png)
![$ a, b$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img73.png)
![$ 0<b</sqrt{p}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img74.png)
![$/displaystyle /left/vert -/frac{r}{p} - /frac{a}{b}/right/vert /leq/frac{1}{b(n+1)} < /frac{1}{b/sqrt{p}}.
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img75.png)
![$/displaystyle c = rb + pa
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img76.png)
![$/displaystyle /vert c/vert < /frac{pb}{b/sqrt{p}} = /frac{p}{/sqrt{p}} = /sqrt{p}
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img77.png)
![$/displaystyle 0 < b^2 + c^2 < 2p.
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img78.png)
![$ c /equiv rb/pmod{p}$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img79.png)
![$/displaystyle b^2 + c^2 /equiv b^2 + r^2 b^2 /equiv b^2(1+r^2) /equiv 0/pmod{p}.
$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img80.png)
![$ b^2 + c^2 = p$](http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/img81.png)