For natural $n$, let $1=d_1I got a nice problem.

Let $n$ be a natural number and $1=d_1<d_2<\dots<d_k=n$ be the positive divisors of $n$. Now, find minimum $n$ such that $2n=d_5^2+d_6^2-1$.

I proceeded as follows. 


Conclusions: The only possibility is $p=2$ and $q=2^4+1=17$. Hence $n=2^4\cdot 17=272$.

Although I have verified, I would like to know if any elegant or shorter solution is there, by that I mean, I am not sure whether this is the smallest number.
I assumed some things right in the beginning, and that is why it is a little unelegant.
I would like to know if there is any better approach to this problem.
Thanks in advance!
 A: Note that $2d_6^2>d_5^2+d_6^2-1=2n$ so $d_6>\sqrt{n}$. Thus, $6>\left\lceil\frac{k}{2}\right\rceil$ so $k\leq 10$.
Similarly, $2d_5^2<d_5^2+d_6^2-1=2n$ so $d_5<\sqrt{n}$. Thus, $5\leq\left\lfloor\frac{k}{2}\right\rfloor$ so $k\geq 9$.
If $k=9$ then $d_5=\sqrt{n}$ which is absurd. ($d_5<\sqrt{n}$)
Hence $k=10$ only. Then $d_5d_6=n$ so $2d_5d_6=d_5^2+d_6^2-1$.
$$(d_5-d_6)^2=1\implies d_6=d_5+1$$
This mean that $n$ is not in the form of $p^9$. Hence, $n=pq^4$ where $p,q$ are prime numebers.
Case 1: $p<q$
The ten divisor of $n$ from smallest to greatest are
$$1,p,q,pq,q^2,pq^2,q^3,pq^3,q^4,pq^4.$$
Thus $pq^2+1=q^3$ which is impossible since $q\mid 1$.
Case 2: $p>q$
Note that since $d_5d_6=d_5(d_5+1)= n$, we have $2\mid n$ so $q=2$.
We can easily check the case $p=3,5,7,11,13$ that they don't work. Thus, $p\geq 17>16$.
The smallest six divisors of $n$ are $1,2,4,8,16,p$. Thus, $p=17$ which works.
Hence, the only possible value of $n$ is $16\cdot 17=272$.
A: Yes, your result is correct and by brute force I confirm that $n=272=2^4\cdot 17$ is the minimum. Your proof is fine to me and it is also elegant. Actually it seems to me that you proved that $272$ is the only possible solution of $2n=d_5^2+d_6^2-1$ and therefore it is the minimum one. 
