Is a way to find all positive integer solution for a linear equation that also satisfy the sum of squares of the unknowns, such that:

Question

Starting from a set of integers: $$\begin{cases} x+y+z+v+w = a+b+c+d+e\\ x^2+y^2+z^2+v^2+w^2 = a^2+b^2+c^2+d^2+e^2 \end{cases}$$ Given that $x, y, z, v, w$ are all positive integers, how can I find if there are another set of integers that satisfy both equations?

For example I know that when $x=271$, $y=106$, $z=438$, $v=385$, $w=42$, $$ \begin{cases} x+y+z+v+w = 1242\\ x^2+y^2+z^2+v^2+w^2 = 426510 \end{cases}$$ Is there a way to find if there are another set of integers that satisfy both equations?

Note: I'm looking for a method quicker than brute-force.

Answer 1

I don't understand why they ask such questions. $$x_1+x_2+...+x_n=y_1+y_2+...+y_k$$ $$x_1^2+x_2^2+...+x_n^2=y_1^2+y_2^2+...+y_k^2$$

When solving such systems, it is always necessary to get rid of the linear equation. $$x_1=y_1+y_2+...+y_k-x_2-x_3-...-x_n$$

After substituting into the second equation, we get a square Diophantine equation. The parameterization of which is written simply. The same principle allows you to solve more complex options.

Nontrivial integer solutions of $\sum_{i=1}^3 a_i ^3=\sum_{i=1}^3 b_i ^3$ and $\sum_{i=1}^3 a_i =\sum_{i=1}^3 b_i$

Finding an answer to Diophantine equations below

https://artofproblemsolving.com/community/c3046h1057888_one_system_of_diophantine_equations

https://artofproblemsolving.com/community/c3046h1053878_the_system_of_equations_quotbquot

https://artofproblemsolving.com/community/c3046h1053831_the_system_of_equations_quotaquot

Answer 2

Let $[(a1,a2,a3,a4,a5),(b1,b2,b3,b4,b5)]$ is a known solution.
Then
$[(t+a1,t+a2,t+a3,t+a4,t+a5),(t+b1,t+b2,t+b3,t+b4,t+b5)]$
is also a solution with any t.

Example:

$41+ 30+ 24+ 12+ 6 = 42+ 30+ 18+ 18+ 5$
$41^2+ 30^2+ 24^2+ 12^2+ 6^2 = 42^2+ 30^2+ 18^2+ 18^2+ 5^2$

Using above one known solution, we can get a new solution as follows.
If we take $t = 100$, we get
$141+ 130+ 124+ 112+ 106 = 142+ 130+ 118+ 118+ 105$
$141^2+ 130^2+ 124^2+ 112^2+ 106^2 = 142^2+ 130^2+ 118^2+ 118^2+ 105^2$

In this way, we can get infinitely many positive integer solutions.