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Cauchy’s Lemma. Let $a$ and $b$ be odd positive integers satisfying $b^2\lt4a$ and $3a \lt b^2+2b+4$. Then, there exist non-negative integers $s,t,u,v$ such that $a=s^2+t^2+u^2+v^2$ and $b=s+t+u+v$.

Corollary. If $a = s^2 + t^2 + u^2 + v^2$ is an odd positive integer, then there exists an odd positive integer $b$ such that $b = s + t + u + v$ and $\sqrt{3a-1}-1\lt b \lt 2\sqrt{a}$.

I got one of the four squares representations of $a = 1387 = 1^2 + 1^2 + 4^2 + 37^2$ and $b = 1 + 1 + 4 + 37 = 43$. But, $b=43$ lies outside of the range $(\sqrt{3\cdot 1387 - 1} - 1, 2\sqrt{1387}) = (63.498,74.485)$ given by the corollary. So, Cauchy's lemma only talks about the existence of a subset of quadruples $(s,t,u,v)$ that are four square representations of $a$. As the example shows, there are other representations for $a$ that yield a $b$ that lies outside the range.

The representation $1387 = 1^2 + 19^2 + 20^2 + 25^2$ gives $b = 1+19+20+25 = 65$ which does lie in the range given by the corollary.

Assuming we start with a four square representation of $a = s^2 + t^2 + u^2 + v^2$ that satisfies the range conditions for $b = s+t+u+v$ given in the corollary and we are looking for

$$a = \overbrace{(p_1^2 + p_2^2 + p_3^2 + p_4^2)}^{p}\overbrace{(q_1^2 + q_2^2 + q_3^2 + q_4^2)}^{q}$$

WLOG take $p \le q$ and $p_1 \le p_2 \le p_3 \le p_4$ and $q_1 \le q_2 \le q_3 \le q_4$.

We have $p \le \sqrt{a} \le q \le a$. Therefore,

$$ \begin{align} p_1^2 + p_2^2 + p_3^2 + p_4^2 \le \sqrt{a} \\ \implies p_4^2 \le \sqrt{a} \\ \implies p_4 \le \sqrt[4]{a} \end{align} $$

Is it possible to constrain $p_1, p_2, p_3, p_4$ (or equivalently $q_1, q_2, q_3, q_4$) using $b=s+t+u+v$ and sharpen this bound? The goal is to generate a four square representation of $p$ or $q$ from the four square representation of $a$ given by $(s,t,u,v)$.

vvg
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    a four square representation can be transformed into another however there is no proof that this can be done for every 4-square representation. Some examples of going from one representation to another are given into this post and some comments. https://math.stackexchange.com/questions/2337259/are-the-4-square-representations-of-a-sum-of-4-squares-of-an-integer-equivalent – user25406 Aug 06 '23 at 12:58
  • I found another 4-square rep that lies outside the range. $1387=36^2 + 9^2 +3^2 +1^2$. I started with $37^2=(1+36)^2$, then you expand and collect numbers into squares leaving $36^2$ and $1^2$ since $2136 + 4^2 +1^2 +1^2=90=9^2 +3^2$. Note that the sum $36+9+3+1=49$. So the sum is getting larger so at one point, the total will be within the range. So I suggest you continue to transform $37$ into $37=36+1=35+2=34+3=...$ until you get the first representation that will be within the range. This work is better done with code (which I can't do). – user25406 Aug 07 '23 at 14:20
  • The problem is not generating the 4 square representations of $n$ that satisfy Cauchy's lemma's range condition. The issue is trying to constrain the 4 square representations of the factors of $n$ based on the four squares representation of $n$ and then applying Euler's identity to further constrain the factor possibilities. – vvg Aug 07 '23 at 17:08
  • Sorry, I misunderstood the question. I did some work on factoring using the 4-square representations here: https://math.stackexchange.com/questions/3088664/can-factoring-with-the-sum-of-4-squares-be-made-more-efficient – user25406 Aug 07 '23 at 17:50

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