TITLE: Find the lowest value of $n$, such that the equation $p^{2}+pq+q^{2}=n$ has only five solutions
QUESTION [11 upvotes]: Let $1\le p<q$ natural numbers such that $p^{2}+pq+q^{2}=n$
Find the lowest value of $n$ so that the equation has only five solutions
is very hard

Notes edited in by MF  
If $\eta(n)$ is the number of $p,q>p$ pairs such that $n=p^2+pq+q^2$, the (lowest) numbers  $\alpha_k: \eta(\alpha_k) = k$ are fascinating:
The first number with $\eta(n) = 4$ is $1729$, which is interesting for other reasons, and is divisible by $13$.  In fact, all lowest numbers $\alpha_k$ for $k>1$ apparently are divisible by $13$:
$$
\eta(91) = 2\\
\eta(637) = 3\\
\eta(1729) = 4\\
\eta(\mbox{answer to the question}) = 5\\
\eta(12103) = 6\\
\eta(53599) = 8
$$
The problem is particularly nice since $a_5 > \alpha_6$
An even harder problem is to determine $\alpha_7$ which is more than $266000$ even though $\alpha_8 < 53600$.

REPLY [4 votes]: All the $\alpha_k$ (the smallest integers with $k$ ways to decompose as 
$p^2+pq+q^2$ that I have been able to find are divisible by $13$. The answer to the problem is $\alpha_5 = 31213 = 7^4\cdot 13$.  The  $\alpha_k$ seem to love prime factors of the form $6r+1$ and hate prime factors of the form $6r-1$.
$$\begin{array}{rl} 
\alpha_2 = & 91 = 7 \cdot 13 \\ 
\alpha_3 = & 637 = 7^2 \cdot 13 \\ 
\alpha_4 = & 1729 = 7 \cdot 13 \cdot 19 \\ 
\alpha_5 = & 31213 = 7^4 \cdot 13 \\ 
\alpha_6 = & 12103 = 7^2 \cdot 13 \cdot 19\\ 
\alpha_7 = & 405769 = 7^4 \cdot 13^2 \\ 
\alpha_8 = & 53599 = 7 \cdot 13 \cdot 19 \cdot 31\\ 
\alpha_9 = & 157339 =7^2 \cdot 13^2 \cdot 19 \\ 
\alpha_{10} = & 593047 = 7^4 \cdot 13 \cdot 19\\  
\alpha_{11} = & 7^{10}×13 (**)\\
\alpha_{12} = & 375193 = 7^2 \cdot 13 \cdot 19 \cdot 31\\  
\alpha_{13} = & 2989441 = 7^2 \cdot 13^2 \cdot 19^2 \\  
\alpha_{14} = & 29059303 = 7^6 \cdot 13 \cdot 19 \\  
\alpha_{15} = & 7709611 = 7^4 \cdot 13^2 \cdot 19 \\  
\alpha_{16} = & 1983163 = 7 \cdot 13 \cdot 19 \cdot 31 \cdot 37\\  
\alpha_{17} = & 7^6×13^4 (**)\\  
\alpha_{18} = & 4877509 = 7^2 \cdot 13^2 \cdot 19 \cdot 31 \\  
\alpha_{19} = & 7^{12}×13^2 (**)\\  
\alpha_{20} = & 18384457 = 7^4 \cdot 13 \cdot 19 \cdot 31 \\  
\alpha_{21} = & 377770939 = 7^6 \cdot 13^2 \cdot 19  \\  
\alpha_{22} = & 146482609 = 7^4 \cdot 13^2 \cdot 19^2  \\  
\alpha_{24} = & 13882141 = 7^2 \cdot 13 \cdot 19 \cdot 31 \cdot 37\\  
\alpha_{27} = & 92671671 = 7^2 \cdot 13^2 \cdot 19^2 \cdot 31 \\  
\end{array}
$$
** - expected result based on prime factorizations.
The value given now for $\alpha_9$ lower than that quoted originally; it was verified as having nine positive solutions by exhaustive search.  With three distinct prime factor $2^{3-1}=4$ of the nine solutions are primitive, the rest nonprimitive.