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:
| \(\ds z_1\) | \(=\) | \(\ds x_1 + y_1\) | ||||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \alpha x_3 + \beta x_4 + \alpha y_3 + \beta y_4\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \alpha \paren {x_3 + y_3} + \beta \paren {x_4 + y_4}\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \alpha z_3 + \beta z_4\) | \(\ds \pmod p\) | |||||||||||
| \(\ds z_2\) | \(=\) | \(\ds x_2 + y_2\) | ||||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \beta x_3 - \alpha x_4 + \beta y_3 - \alpha y_4\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \beta \paren {x_3 + y_3} - \alpha \paren {x_4 + y_4}\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \beta z_3 - \alpha z_4\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \vec x + \vec y\) | \(=\) | \(\ds \vec z\) | |||||||||||
| \(\ds \) | \(\in\) | \(\ds L\) |
So $L$ is closed under vector addition.
| \(\ds -x_1\) | \(\equiv\) | \(\ds -\paren {\alpha x_3 + \beta x_4}\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \alpha \paren {-x_3} + \beta \paren {-x_4}\) | \(\ds \pmod p\) | |||||||||||
| \(\ds -x_2\) | \(\equiv\) | \(\ds -\paren {\beta x_3 - \alpha x_4}\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \beta \paren {-x_3} - \alpha \paren {-x_4}\) | \(\ds \pmod p\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -\vec x\) | \(\in\) | \(\ds L\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds L\) | \(\le\) | \(\ds \R^4\) |
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:
| \(\ds \alpha c_1 + \beta c_2 + p c_3\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \beta c_1 - \alpha c_2 + p c_4\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds c_1\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds c_2\) | \(=\) | \(\ds 0\) |
and so:
| \(\ds \alpha c_1 + \beta c_2\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \alpha c_1 - \beta c_2\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds c_3\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds c_4\) | \(=\) | \(\ds 0\) |
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:
| \(\ds \map {\dim_\R} {\span_\R L}\) | \(=\) | \(\ds \dim_\Z L\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \card {\set {\tuple {\alpha, \beta, 1, 0}, \tuple {\beta, -\alpha, 0, 1}, \tuple {p, 0, 0, 0}, \tuple {0, p, 0, 0} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 4\) |
Thus:
- $\span_\R L = \R^4$
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:
- $\map \det L = \# \paren {\Z^4 / L} < p^2$
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Let $\norm {\, \cdot \,}$ denote the Euclidean metric.
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Consider $\map {B_{\sqrt {2 p} } } {\vec 0}$, the open ball of radius $\sqrt {2 p}$.
Then:
- $\forall \vec x, \vec y \in \map {B_{\sqrt {2 p} } } {\vec 0}, \forall t \in \closedint 0 1$:
| \(\ds \norm {t \vec x + \paren {1 - t} \vec y}\) | \(\le\) | \(\ds \norm {t \vec x} + \norm {\paren {1 - t} \vec y}\) | Definition of Metric | |||||||||||
| \(\ds \) | \(=\) | \(\ds \size t \norm {\vec x} + \size {1 - t} \norm {\vec y}\) | Definition of Metric | |||||||||||
| \(\ds \) | \(<\) | \(\ds \size t \sqrt {2 p} + \size {1 - t} \sqrt {2 p}\) | Definition of Open Ball $\map {B_{\sqrt{2 p} } } {\vec 0}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\size t + \size {1 - t} } \sqrt {2 p}\) | Real Multiplication Distributes over Addition | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {t + 1 - t} \sqrt{2 p}\) | since $t \in \closedint 0 1 \implies t \ge 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt {2 p}\) | 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 $\R^n$):
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- $\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } = \dfrac {\pi^2 \paren {\sqrt {2 p} }^4} 2 = 2 \pi^2 p^2$
Since $2\pi^2 > 1$:
- $\map {\operatorname {Vol} } {\map {B_{\sqrt{2 p} } } {\vec 0} } > \map \det L$
- $L \cap \map {B_{\sqrt{2 p} } } {\vec 0} \ne \O$
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:
| \(\ds a_1^2 + a_2^2 + a_3^2 + a_4^2\) | \(\equiv\) | \(\ds \paren {\alpha a_3 + \beta a_4}^2 + \paren {\beta a_3 - \alpha a_4}^2 + a_3^2 + a_4^2\) | \(\ds \pmod p\) | Definition of $L$ | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \alpha^2 a_3^2 + 2 \alpha \beta a_3 a_4 + \beta^2 a_4^2 + \beta^2 a_3^2 - 2 \alpha \beta a_3 a_4 + \alpha^2 a_4^2 + a_3^2 + a_4^2\) | \(\ds \pmod p\) | Binomial Theorem | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \paren {\alpha^2 + \beta^2 + 1} \paren {a_3 + a_4}\) | \(\ds \pmod p\) | Real Multiplication Distributes over Addition | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds 0 \paren {a_3 + a_4}\) | \(\ds \pmod p\) | by $(1)$ from above | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod p\) |
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 $x$ or $y$ being primes.
Suppose either one is equal to $1$.
Then $x * 1 = x$ is a sum of four squares.
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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 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 Fundamental Theorem of Arithmetic, every number can be written as a unique product of primes.
Then
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