Cauchy's Lemma (Number Theory)

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

Suppose $a$ and $b$ satisfy:
 * $b^2 < 4a$
 * $3 a < b^2 + 2 b + 4$

Then there exist nonnegative integers $s, t, u, v$ such that:

Proof
Since $a$ is odd, we can write $a = 2 k + 1$ for some positive integers $k$.

Then:

In the proof of Integer is Sum of Three Triangular Numbers, it is shown that there exists $3$ odd integers $x, y, z$ such that:
 * $4 a - b^2 = x^2 + y^2 + z^2$

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

One of $b + x + y \pm z$ must be divisible by $4$.

To see this, we assume that $b + x + y + z$ is not divisible by $4$.

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


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

Writing $z = 2 l + 1$:

So $b + x + y - z$ is divisible by $4$.

Choosing the case when $b + x + y \pm z$ is divisible by $4$, we define:

We claim that $s, t, u, v$ are nonnegative, and will satisfy:

First we check that $s, t, u, v$ are nonnegative.

Since $x, y, z$ 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:


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

Now:

showing that $s, t, u, v$ are nonnegative.

Now we check $(1)$:

Now we check $(2)$: