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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 $4$.

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.

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.

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 $4$.

Fix a positive integer $r$. Is it possible to solve the system of equations given by:

$$\begin{equation}\sum_{1\leq i\leq r}X_i^2\equiv0\pmod{X_k}(1\leq i\leq k)\end{equation}$$

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.

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.