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:

Proof
Because $a$ is odd, we can write:
 * $a = 2 k + 1$

for some positive integer $k$.

Then:

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 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 $z = 2 l + 1$:

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

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

We define:

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

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:

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

Now we check $(1)$:

Now we check $(2)$:

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