Finding integer zeroes for a particular family of equations [closed]

Question

Given $p,q\in\mathbb Z^+$, and a vector $v=(x_1,\dots,x_{p+q})$ we consider the function $\chi(v)$:

$$\chi(v)=x_1^2+\dots+x_p^2-x_{p+1}^2-\dots-x_{p+q}^2$$

We wish to find solutions to $\chi(v)=0$ for all $p,q$ with the condition that $x_i\in\mathbb Z^+$. This has no solutions if $p=0$ or $q=0$.

I have found the earliest solutions for $p+q=50$ where valid here.

There are two questions:

1. For any $p\ne0\land q\ne0$ are there solutions?
2. Is there a simple method to find solutions for any valid $p,q$?

Answer 1

Here is an easy argument, answering affirmatively both questions. I have interpreted the slightly ambiguous wording as meaning that for each choice of $p,q$, with both positive (which means strictly positive), there exists a solution with all entries (strictly) positive.

Lemma. Given $q \geq 1$, there exists a tuple of positive integers, $(a; b_1, b_2, \dots, b_q)$ such that $a $ is odd and $a^2 = \sum b_i^2$.

Proof. By induction. If $q = 1$, set $a = b_1 =1$. Otherwise assume true for $q-1$; there exists a strictly positive tuple, $(r; b_1, b_2, \dots, b_{q-1}) $, such that $r^2 = \sum_{i=1}^{q-1} b_i^2$ with odd $r > 1 $. Then we can write $r^2 = ((r^2+1)/2)^2 - ((r^2 -1)/2)^2 =e^2 -f^2$, as a difference of square integers, with $e$ odd, and $f$ even but nonzero. Set $a = e$ (so $a$ is odd) and $b_q = f$. Then $$\eqalign{ a^2 - \sum_1^q b_i^2 &= \left(r^2 - \sum_{i=1}^{q-1} b_i^2\right) + a^2 - r^2 - b_q^2\cr & = 0 + 0 = 0.\cr }$$

[Hypothesis about oddness is needed for the differences of squares argument.]

Corollary. If $p, q$ are positive integers, there exists a strictly positive integer solution $(x_1, \dots, x_p; y_1, \dots y_q) $ to the equation $$ \sum_{j=1}^p x_j^2 = \sum_{i=1}^q y_i^2. $$

Proof. Without loss of generality, $p \leq q$. If $p= q$, set all the variables equal to $1$. Otherwise, $p < q$; by setting $x_1 = x_2 = \dots= x_{p-1} = 1 = y_1 = y_2 = \dots = y_{p-1}$, we immediately reduce to the case that $p = 1$ and $q > 1$. But this is the conclusion of the lemma.

Presumably, by actually applying the induction argument with the smallest possible choices for $r$, one obtains small (or even the smallest) solutions (where we measure smallness by the sum)?

The problem would be more difficult if all the entries of the solution were required to be distinct as well.

Answer 2

This is an old question, and there will probably be better answers, but in the meantime, check out Browning and Dietmann's paper, and (our own) Paul Garrett's notes.

Answer 3

$$\sum_{i=1}^{p-1}x_i^2+\frac{1}{4}\bigg(\sum_{i=1}^{p-1}x_i^2-\sum_{i=p+1}^{p+q-1}x_i^2-1 \Bigg)^2=\sum_{i=p+1}^{p+q-1}x_i^2+\frac{1}{4}\bigg(\sum_{i=1}^{p-1}x_i^2-\sum_{i=p+1}^{p+q-1}x_i^2+1 \Bigg)^2\ ,$$ so you can e.g. choose freely $x_1,\dots,x_{p-1},x_{p+1},\dots,x_{p+q-1}$ with odd sum, and define correspondingly $x_p$ and $x_{p+q}$.If $x_1,\dots,x_{p-1},x_{p+1},\dots,x_{p+q-1}$ are distinct, and if $\sum_{i=1}^{p-1}x_i^2$ is suitably larger than $\sum_{i=p+1}^{p+q-1}x_i^2 $, then $x_p$ and $x_{p+q}$ will be also distinct form each other and from the previously chosen $x_i$.

Answer 4

Here is another approach.

(1,2) and (2,1) are pythagorean triples and are parametrized.

Set $x_1=x,x_{p+1}=x-1$. Then $x_1^2-x_{p+1}^2=2x-1$.

$2x-1$ represents all positive odd integers, so it is difference of two squares and sum of two squares infinitely often.

If $p>1$, set all variables other than $x,x_2,x_{2+p}$ to $4$.

For integer $m$, we have $2x-1+4m$ difference of two squares and this has infinitely many solutions.

$$ 2x-1+4m + x_2^2-x_{2+p}^2=0$$

If $p=1$, set all variables other than $x,x_{2+p},x_{3+p}$ to $4$.

For integer $m$, we have $2x-1+4m$ sum of two squares and this has infinitely many solutions, since it is prime of the form $4k+1$ infinitely often by primes in arithmetic progressions.

$$2x-1+4m-x_{2+p}^2-x_{3+p}^2=0$$

To avoid factorization in sum of two squares, find prime in the arithmetic progression.