A problem on a specific integer partition Let $n$ be a positive integer, we consider partitions of the following form :
$$n = d^{2}_{1} + d^{2}_{2} + ... + d^{2}_{r}$$ such that : 


*

*$d_{i}\vert n$

*$1=d_{1}<d_{2} \le d_{3} \le ... \le d_{r}$  

*$gcd(d_{2},d_{3},...,d_{r}) = 1$     


We call $r$ the rank of the partition.
Examples :  


*

*$60 = 1+3^{2}+3^{2}+4^{2}+5^{2}$ 

*$168 = 1+3^{2}+3^{2}+6^{2}+7^{2}+8^{2} $  


Remark : Let $G$ be a non-cyclic finite simple group, and $(H_{i})_{i}$ its irreducible representations, then : $$ord(G) = \sum_{i} dim(H_{i})^{2}$$ is an example. The first example comes from the group $A_{5}$, and the second, from the group $A_{1}(7)$.
Of course, there are examples not coming from a group : $60 = 1 + 3^{2} + 5^{2}+5^{2}$ 

Problem : Are there only finitely many such partitions for a fixed rank $r$ ?     

Motivation : The study of finite quantum groups (see here).
Final remark: Let $p_{r}(n)$ be the number of such partitions for $n$, of rank $r$.
If the problem has a positive answer, then $N_{r} = max\{n : p_{r}(n) \ne 0   \} < \infty$
In this case, the generating function $P_{r}(x) = \sum_{n}p_{r}(n)x^{n}$, is a polynomial of degree $N_{r}$.
Bonus problem : Find an evaluation of the number $N_{r}$.     
 A: Update: Sebastien and I just found out that the equation $X_1^2+\dots+X_r^2=mX_1\dots X_r$ which evolved in the comments below is a classical topic, named the Hurwitz equation. See this encyclopedia entry. It was actually discussed on MO before, see this post. So most of the comments below are obsolete.
Existence: For $r=3$ there are infinitely many such partitions. Let $F_i$ be the $i$th Fibonacci number. Then:
\begin{equation}
1+F_{2i-1}^2+F_{2i+1}^2=3F_{2i-1}F_{2i+1}.
\end{equation}
This follows from Cassini's identity $F_{n-1}F_{n+1}-F_n^2=(-1)^n$ upon replacing $F_n$ with $F_{n+1}-F_{n-1}$, and setting $n=2i$.
Uniqueness: We now show the following:
 Let $1\le b\le a$ be integers such that $abm=a^2+b^2+1$ for an integer $m$. Then $m=3$ and either $a=b=1$, or $b=F_{2i-1}$, $a=F_{2i+1}$.
If $a=b$, then of course $a=b=1$ and $m=3$. Suppose now that $b<a$. Set $c=mb-a$. Then
\begin{align}
c^2+b^2+1 &=
b^2m^2-2abm+a^2+b^2+1\\
&= b^2m^2-2abm +mab\\
&= mb(bm-a)\\
&= mbc
\end{align}
So the transformation $(a,b)\mapsto (b,mb-a)=(b,c)$ sends $(a,b)$ to another solution $(b,c)$. Note that $c>0$ as $b>0$. From
\begin{equation}
ca = (mb-a)a = b^2+1 \le b(b+1) \le ba
\end{equation}
we obtain $c\lt b$, unless $b=1$ and $a=2$. So by descend we finally arrive at $(a,b)=(2,1)$, hence $m=3$.
We see that $m=3$, and that every solution of $a^2+b^2+1=3ab$ with $a>b$ has the form $(a_k,a_{k-1})$, where $(a_k)$ is the sequence defined by $a_1=1$, $a_2=2$ and $a_k=3a_{k-1}-a_{k-2}$. However, the sequence $F_{2i+1}$ fulfills this recurence, and the assertion follows.
A: Since the question started with examples coming from degrees of irreducible characters of groups, it is perhaps worth noting that if we limit ourselves to group examples, there are only finitely many of a given rank. Since the "rank" is the number of irreducible characters of the group, and that is equal to the number of conjugacy classes, an old result of Landau applies. It asserts that there is some function $L(r)$ such that if a 
$~finite~$ group $G$ has exactly $r$ classes, then $|G| \le L(r)$. (Note that the finiteness hypothesis is essential.) Landau's theorem appears as Theorem 4.13 of my algebra text (Algebra: A Graduate Course).
