# Cauchy's Lemma (Number Theory)

## Theorem

Let $a$ and $b$ be odd positive integers.

Suppose $a$ and $b$ satisfy:

$b^2 < 4 a$
$3 a < b^2 + 2 b + 4$

Then there exist non-negative integers $s, t, u, v$ such that:

 $\ds a$ $=$ $\ds s^2 + t^2 + u^2 + v^2$ $\ds b$ $=$ $\ds s + t + u + v$

## Proof

Because $a$ is odd, we can write:

$a = 2 k + 1$

for some positive integer $k$.

Then:

 $\ds 4 a - b^2$ $=$ $\ds 4 \paren {2 k + 1} - b^2$ $\ds$ $\equiv$ $\ds 8 k + 4 - 1$ $\ds \pmod 8$ Odd Square Modulo 8 $\ds$ $\equiv$ $\ds 3$ $\ds \pmod 8$

Because $b^2 < 4 a$, we have that $4 a - b^2$ is a positive integer.

By Integer as Sum of Three Odd Squares, there exist $3$ odd positive integers $x, y, z$ such that:

$4 a - b^2 = x^2 + y^2 + z^2$

Because $b, x, y, z$ are all odd integers, $b + x + y + z$ must be even.

It is now to be shown that $b + x + y \pm z$ is divisible by $4$.

Suppose that $b + x + y + z$ is not divisible by $4$.

Because $b + x + y + z$ is even:

$b + x + y + z \equiv 2 \pmod 4$

Writing $z = 2 l + 1$:

 $\ds b + x + y - z$ $=$ $\ds b + x + y + z - 2 z$ $\ds$ $\equiv$ $\ds 2 - 2 \paren {2 l + 1}$ $\ds \pmod 4$ $\ds$ $\equiv$ $\ds 2 - 4 l - 2$ $\ds \pmod 4$ $\ds$ $\equiv$ $\ds 0$ $\ds \pmod 4$

That is:

if $b + x + y + z$ is not divisible by $4$
then $b + x + y - z$ is divisible by $4$.

$\Box$

Let us choose the case such that $b + x + y \pm z$ is divisible by $4$.

We define:

 $\ds s$ $=$ $\, \ds \frac {b + x + y \pm z} 4 \,$ $\ds$ $\ds t$ $=$ $\, \ds \frac {b + x - y \mp z} 4 \,$ $\, \ds = \,$ $\ds \frac {b + x} 2 - s$ $\ds u$ $=$ $\, \ds \frac {b - x + y \mp z} 4 \,$ $\, \ds = \,$ $\ds \frac {b + y} 2 - s$ $\ds v$ $=$ $\, \ds \frac {b - x - y \pm z} 4 \,$ $\, \ds = \,$ $\ds \frac {b \pm z} 2 - s$

We are to show that $s, t, u, v$ are non-negative, and will satisfy:

 $\text {(1)}: \quad$ $\ds a$ $=$ $\ds s^2 + t^2 + u^2 + v^2$ $\text {(2)}: \quad$ $\ds b$ $=$ $\ds s + t + u + v$

First we show that $s, t, u, v$ are non-negative.

Because $x, y, z$ are positive:

$s, t, u, v \ge \dfrac {b - x - y - z} 4$

So we need to show:

$\dfrac {b - x - y - z} 4 \ge 0$

or equivalently:

$\dfrac {b - x - y - z} 4 > -1$

Now:

 $\ds x + y + z$ $\le$ $\ds \sqrt {3 \paren {x^2 + y^2 + z^2} }$ Inequality of Hölder Means $\ds$ $=$ $\ds \sqrt {3 \paren {4 a - b^2} }$ $4 a - b^2 = x^2 + y^2 + z^2$ $\ds$ $<$ $\ds \sqrt {4 \paren {b^2 + 2 b + 4} - 3 b^2}$ $3 a < b^2 + 2 b + 4$ $\ds$ $=$ $\ds \sqrt {b^2 + 8 b + 16}$ $\ds$ $=$ $\ds b + 4$ Square of Sum $\ds \leadsto \ \$ $\ds \frac {b - x - y - z} 4$ $>$ $\ds -1$

showing that $s, t, u, v$ are non-negative.

Now we check $(1)$:

 $\ds s^2 + t^2 + u^2 + v^2$ $=$ $\ds \paren {\frac {b + x + y \pm z} 4}^2 + \paren {\frac {b + x - y \mp z} 4}^2 + \paren {\frac {b - x + y \mp z} 4}^2 + \paren {\frac {b - x - y \pm z} 4}^2$ $\ds$ $=$ $\ds \frac 1 {16} \leftparen {b^2 + x^2 + y^2 + z^2 + 2 b x + 2 b y \pm 2 b z + 2 x y \pm 2 x z \pm 2 y z}$ $\ds$  $\ds + \quad b^2 + x^2 + y^2 + z^2 + 2 b x - 2 b y \mp 2 b z - 2 x y \mp 2 x z \pm 2 y z$ $\ds$  $\ds + \quad b^2 + x^2 + y^2 + z^2 - 2 b x + 2 b y \mp 2 b z - 2 x y \pm 2 x z \mp 2 y z$ $\ds$  $\ds + \quad \rightparen {b^2 + x^2 + y^2 + z^2 - 2 b x - 2 b y \pm 2 b z + 2 x y \mp 2 x z \mp 2 y z}$ $\ds$ $=$ $\ds \frac 1 {16} \paren {4 \paren {b^2 + x^2 + y^2 + z^2} }$ $\ds$ $=$ $\ds \frac {4 a} 4$ $\ds$ $=$ $\ds a$

Now we check $(2)$:

 $\ds s + t + u + v$ $=$ $\ds \frac {b + x + y \pm z} 4 + \frac {b + x - y \mp z} 4 + \frac {b - x + y \mp z} 4 + \frac {b - x - y \pm z} 4$ $\ds$ $=$ $\ds \frac {4 b} 4 + \frac {2 x - 2 x} 4 + \frac {2 y - 2 y} 4 \pm \frac {2 z - 2 z} 4$ $\ds$ $=$ $\ds b$

Therefore $s, t, u, v$ as defined are the integers we are looking for.

$\blacksquare$

## Source of Name

This entry was named for Augustin Louis Cauchy.