Fix a positive integer $r$. Describe the solutions to the system of equations given by:


$$\begin{equation}\sum\_{1\leq i\leq r}X\_i^2\equiv0\pmod{X\_k}(1\leq k\leq r)\end{equation}$$


**Example:** In the case that $r=3$, $(X\_1,X\_2,X\_3)=(1,5,13)$ is a solution, because $1^2+5^2+13^2=195=3\cdot 5\cdot 13$ is divisible by $1$, $5$ and $13$.


**Motivation:** The degrees of the irreducible characters $\text{Irr}(G)$ of a finite group $G$ satisfy the following relation:


$$\begin{equation}\sum\_{\chi\in\text{Irr}(G)}\chi(1)^2=|G|\end{equation}$$


Furthermore, for every $\chi\in\text{Irr}(G)$, $\chi(1)$ divides $|G|$. So the broader question is: is there some way to arithmetically characterize sequences of positive integers which are the irreducible degrees of some finite group?


The example I gave above does *not* qualify. No group has irreducible degrees $(1,5,13)$. Such a group would have $195$ elements, but since $195$ is a squarefree number, our hypothetical group with $195$ elements would have to be solvable. But nontrivial solvable groups must have nonprincipal linear characters, so $(1,5,13)$ cannot be its set of irreducible degrees.


(27th Aug)
**Example:**


In the case that $r=4$, $(X\_1,X\_2,X\_3,X\_4)=(1,1,2,4)$ is a solution, because $1^2+1^2+2^2+4^2=22=2\cdot 11$. But this does not correspond to the irreducible degrees of some finite group $G$. Suppose it did, so $|G|=22$. Then $G$ must be non-abelian, so $G\cong C\_{11}\rtimes C\_2$. However, every irreducible character of $C\_{11}\rtimes C\_2$ has either degree $1$ or $2$, which contradicts the assumption that $G$ has an irreducible character of degree $4$.



An arithmetic problem involving a system of equations
Even for $r=3$ the question isn't easy. For instance a special case is
[Markov's diophantine equation](https://en.wikipedia.org/wiki/Markov_number) $a^2+b^2+c^2=3abc$, for which one infinite family of solutions is $(1, F\_{2n-1}, F\_{2n+1})$, where $F\_i$ is the $i$th Fibonacci number.


