Lagrange's Four Square Theorem/Proof 2

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Theorem

Every positive integer can be expressed as a sum of four squares.


Proof

Proof for Odd Primes

Suppose $p$ is an odd prime.

Define:

$S := \set {\alpha^2 \pmod p: \alpha \in \hointr 0 {\dfrac p 2} \cap \Z}$

Define:

$S' := \set {-1 - \beta^2 \pmod p: \beta \in \hointr 0 {\dfrac p 2} \cap \Z}$


Suppose for $\alpha, \alpha' \in S$:

$\alpha^2 \equiv \alpha'^2 \pmod p$

Obviously:

$\paren {\alpha + \alpha'} \paren {\alpha - \alpha'} = \alpha^2 - \alpha'^2 \equiv 0 \pmod p$

Since $0 \le \alpha$, $\alpha' < \dfrac p 2$:

$\alpha + \alpha' \not \equiv 0 \pmod p$

Therefore $\alpha - \alpha' \equiv 0 \pmod p$.

This shows that $\alpha = \alpha'$.

Thus we have $\size S = \size {\hointr 0 {\dfrac p 2} \cap \Z} = 1 + \dfrac {p - 1} 2 = \dfrac {p + 1} 2$.


Choose $\beta, \beta' \in S'$:

$-1 - \beta^2 \equiv -1 - \beta'^2 \pmod p$

By simple algebraic manipulation:

$-1 - \beta^2 \equiv -1 - \beta'^2 \pmod p \iff \beta^2 \equiv \beta'^2 \pmod p$

Then by the same reasoning as above:

$\card {S'} = \card S = \dfrac {p + 1} 2$

By the Pigeonhole Principle:

$S \cap S' \ne \O$

Thus $\exists \alpha, \beta \in \Z$:

$(1): \quad \alpha^2 + \beta^2 + 1 \equiv 0 \pmod p$

Define:

$L = \set {\vec x = \tuple {x_1, x_2, x_3, x_4} \in \Z^4: x_1 \equiv \alpha x_3 + \beta x_4 \pmod p, x_2 \equiv \beta x_3 - \alpha x_4 \pmod p}$

If $\vec x$, $\vec y \in L$ and $\vec z = \tuple {z_1, z_2, z_3, z_4}$, then:

\(\displaystyle z_1\) \(=\) \(\displaystyle x_1 + y_1\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \alpha x_3 + \beta x_4 + \alpha y_3 + \beta y_4\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \alpha \paren {x_3 + y_3} + \beta \paren {x_4 + y_4}\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \alpha z_3 + \beta z_4\) \(\displaystyle \pmod p\)
\(\displaystyle z_2\) \(=\) \(\displaystyle x_2 + y_2\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \beta x_3 - \alpha x_4 + \beta y_3 - \alpha y_4\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \beta \paren {x_3 + y_3} - \alpha \paren {x_4 + y_4}\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \beta z_3 - \alpha z_4\) \(\displaystyle \pmod p\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \vec x + \vec y\) \(=\) \(\displaystyle \vec z\)
\(\displaystyle \) \(\in\) \(\displaystyle L\)

So $L$ is closed under vector addition.


\(\displaystyle -x_1\) \(\equiv\) \(\displaystyle -\paren {\alpha x_3 + \beta x_4}\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \alpha \paren {-x_3} + \beta \paren {-x_4}\) \(\displaystyle \pmod p\)
\(\displaystyle -x_2\) \(\equiv\) \(\displaystyle -\paren {\beta x_3 - \alpha x_4}\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle \beta \paren {-x_3} - \alpha \paren {-x_4}\) \(\displaystyle \pmod p\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle -\vec x\) \(\in\) \(\displaystyle L\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle L\) \(\le\) \(\displaystyle \R^4\)

So $L$ has additive inverses.

By Two-Step Subgroup Test, $\struct {L, +}$ is a subgroup of $\struct {\R^4, +}$.


For each $\vec x = \tuple {x_1, x_2, x_3, x_4} \in L$, one can write:

$\vec x = x_3 \tuple {\alpha, \beta, 1, 0} + x_4 \tuple {\beta, -\alpha, 0, 1} + \floor {\dfrac {x_1} p} \tuple {p, 0, 0, 0} + \floor {\dfrac{x_2} p} \tuple {0, p, 0, 0}$

Thus:

$\set {\tuple {\alpha, \beta, 1, 0}, \tuple {\beta, -\alpha, 0, 1}, \tuple {p, 0, 0, 0}, \tuple {0, p, 0, 0} }$

spans $L$.


Suppose we have for some $c_1, c_2, c_3, c_4 \in \Z$:

$\vec 0 = c_1 \tuple {\alpha, \beta, 1, 0} + c_2 \tuple {\beta, -\alpha , 0, 1} + c_3 \tuple {p, 0, 0, 0} + c_4 \tuple {0, p, 0, 0}$


Extracting the various coordinates:

\(\displaystyle \alpha c_1 + \beta c_2 + p c_3\) \(=\) \(\displaystyle 0\)
\(\displaystyle \beta c_1 - \alpha c_2 + p c_4\) \(=\) \(\displaystyle 0\)
\(\displaystyle c_1\) \(=\) \(\displaystyle 0\)
\(\displaystyle c_2\) \(=\) \(\displaystyle 0\)

and so:

\(\displaystyle \alpha c_1 + \beta c_2\) \(=\) \(\displaystyle 0\)
\(\displaystyle \alpha c_1 - \beta c_2\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle c_3\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle c_4\) \(=\) \(\displaystyle 0\)


So:

$\set {\tuple {\alpha, \beta, 1, 0}, \tuple {\beta, -\alpha, 0, 1}, \tuple {p, 0, 0, 0}, \tuple {0, p, 0, 0} }$

is linearly independent and thus a basis for $L$.


Thus:

\(\displaystyle \map {\dim_\R} {\operatorname{span}_\R L}\) \(=\) \(\displaystyle \dim_\Z L\)
\(\displaystyle \) \(=\) \(\displaystyle \card {\set {\tuple {\alpha, \beta, 1, 0}, \tuple {\beta, -\alpha, 0, 1}, \tuple {p, 0, 0, 0}, \tuple {0, p, 0, 0} } }\)
\(\displaystyle \) \(=\) \(\displaystyle 4\)

Thus:

$\operatorname{span}_\R L = \R^4$

So $L$ is a lattice.

Define the quotient map:

$\varphi: \N_p^2 \times \set {\tuple {0, 0} } \to \Z^4 / L$
$\tuple {x, y, 0, 0} \mapsto \sqbrk {\tuple {x, y, 0, 0} }$


Since quotient maps are surjective:

$\paren {\Z^4 / L} \le \size {\N_p^2 \times \set {\tuple {0, 0} } } = p^2$

So:

$\map \det L = \# \paren {\Z^4 / L} < p^2$



Let $\norm {\, \cdot \,}$ denote the Euclidean metric.



Consider $\map {B_{\sqrt {2 p} } } {\vec 0}$, the open ball of radius $\sqrt {2 p}$.

Then:

$\forall \vec x, \vec y \in \map {B_{\sqrt {2 p} } } {\vec 0}, \forall t \in \closedint 0 1$:
\(\displaystyle \norm {t \vec x + \paren {1 - t} \vec y}\) \(\le\) \(\displaystyle \norm {t \vec x} + \norm {\paren {1 - t} \vec y}\) Definition of Metric
\(\displaystyle \) \(=\) \(\displaystyle \size t \norm {\vec x} + \size {1 - t} \norm {\vec y}\) Definition of Metric
\(\displaystyle \) \(<\) \(\displaystyle \size t \sqrt {2 p} + \size {1 - t} \sqrt {2 p}\) Definition of Open Ball $\map {B_{\sqrt{2 p} } } {\vec 0}$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\size t + \size {1 - t} } \sqrt {2 p}\) Real Multiplication Distributes over Addition
\(\displaystyle \) \(=\) \(\displaystyle \paren {t + 1 - t} \sqrt{2 p}\) since $t \in \closedint 0 1 \implies t \ge 0$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {2 p}\) since $t + 1 - t = 1$ is the multiplicative identity of $\R$


Thus the line between $\vec x$ and $\vec y$ is contained in $\map {B_{\sqrt{2 p} } } {\vec 0}$.

So $\map {B_{\sqrt{2 p} } } {\vec 0}$ is convex.

Let $\vec x \in \map {B_{\sqrt{2 p} } } {\vec 0}$.

Then $\vec x < \sqrt {2 p}$.

That means:

$\norm {-\vec x} = \size {-1} \norm {\vec x} = \norm {\vec x} < \sqrt{2 p}$

So:

$-\vec x \in \map {B_{\sqrt{2 p} } } {\vec 0}$

Then $\map {B_{\sqrt{2 p} } } {\vec 0}$ is symmetric about the origin.

By the formula for the volume of an n-ball (with respect to the standard measure on $\R^n$):



$\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } = \dfrac {\pi^2 \paren {\sqrt {2 p} }^4} 2 = 2 \pi^2 p^2$

Since $2\pi^2 > 1$:

$\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } > \map \det L$

By Minkowski's Theorem:

$L \cap \map {B_{\sqrt{2 p} } } {\vec 0} \ne \O$

Thus, if:

$\vec a = \tuple {a_1, a_2, a_3, a_4} \in L \cap \map {B_{\sqrt{2 p} } } {\vec 0}$

then:

$0 < a_1^2 + a_2^2 + a_3^2 + a_4^2 = \norm {\vec a}^2 < 2 p$


However:

\(\displaystyle a_1^2 + a_2^2 + a_3^2 + a_4^2\) \(\equiv\) \(\displaystyle \paren {\alpha a_3 + \beta a_4}^2 + \paren {\beta a_3 - \alpha a_4}^2 + a_3^2 + a_4^2\) \(\displaystyle \pmod p\) Definition of $L$
\(\displaystyle \) \(\equiv\) \(\displaystyle \alpha^2 a_3^2 + 2 \alpha \beta a_3 a_4 + \beta^2 a_4^2 + \beta^2 a_3^2 - 2 \alpha \beta a_3 a_4 + \alpha^2 a_4^2 + a_3^2 + a_4^2\) \(\displaystyle \pmod p\) Binomial Theorem
\(\displaystyle \) \(\equiv\) \(\displaystyle \paren {\alpha^2 + \beta^2 + 1} \paren {a_3 + a_4}\) \(\displaystyle \pmod p\) Real Multiplication Distributes over Addition
\(\displaystyle \) \(\equiv\) \(\displaystyle 0 \paren {a_3 + a_4}\) \(\displaystyle \pmod p\) by $(1)$ from above
\(\displaystyle \) \(\equiv\) \(\displaystyle 0\) \(\displaystyle \pmod p\)

So $\exists k \in \Z$:

$(2): \quad 0 < a_1^2 + a_2^2 + a_3^2 + a_4^2 = k p \le 2 p$

Dividing $(2)$ by $p$:

$0 < k < 2 \implies k = 1$

Thus:

$a_1^2 + a_2^2 + a_3^2 + a_4^2 = p$

$\Box$


Proof for Composites

Suppose $x$, $y \in \Z$ are a sum of four squares with neither of $x$, $y$ being primes.

Suppose one of them is equal to $1$.

Then $x * 1 = x$ is a sum of four squares.



Suppose neither of them is equal to $1$.

Let $\mathbb H$ denote the set of Hurwitz quaternions.

Let $N: \mathbb H \to \R, a + b i + c j + d k \mapsto a^2 + b^2 + c^2 + d^2$ be the standard norm on $\mathbb H$.

Notice:

$x$ is a sum of four squares if and only if $x$ is a norm of a Hurwitz quaternion

Then:

$\exists \mu, \lambda \in \mathbb H: x = \map N \mu, y = \map N \lambda$

From Norm is a Homomorphism:

$ x y = \map N \mu \, \map N \lambda = \map N {\mu \lambda}$

Since $\mathbb H$ is a ring, we have:

$\mu \lambda \in \mathbb H$

and thus $x y$ is a sum of four squares.

From the Unique Factorization Theorem, every number can be written as a unique product of primes.

Then