Let $1=This problem is stated as follows:

Let $1=a_1<a_2<a_3<...<a_k=n$ be the divisors of a positive integer $n$. Find all $n$ such that $n={a_2}^2+{a_3}^3$

I have some issues understanding this problem and would like to see some approaches to it.
 A: The only such $n$ is $68$
To see this, note that $a_2=p$ must be the least prime that divides $n$.  But then $$n=p^2+a_3^3\implies p\,|\,a_3\implies a_3=p^2$$
But in that case we must have $$n=p^2+p^6=p^2(1+p^4)$$
Now, if $p$ were odd then $1+p^4$ would be even, whence $2\,|\,n$ contradicting the minimal property of $p$.  Thus $p=2$ and $$n=4+64=68$$
A: $n$ cannot be prime because there would not be a value for $a_3$.  $a_2$ has to be the smallest prime dividing $n$.  $a_3$ can either be the second smallest prime dividing $n$ or $a_2^2$.  We can look at each case.
In the first case, call the primes $p,q$.  Then we are given $n=p^2+q^3$.  If $p,q$ are coprime, $n$ will be coprime to both and they cannot be factors of $n$ so this case is impossible.
In the second case, call the prime $p$ and we are given $n=p^2+p^6$.  Any other prime dividing $n$ must be greater than $p^2$.  If $p$ is odd, $n$ is even and has a factor $2$ which is less than $p$.  The only solution is $p=2, a_2=2,a_3=4,n=68=2^2\cdot 17$ and $17 \gt 2^2$
A: $a_2$ must be prime and then from $a_2\mid n-a_2^2=a_3^3$, we conclude that $a_2\mid a_3$. This is only possible if $a_3=a_2^2$. Then $n=p^2+p^3=p^2(p+1)$ for some prime $p$. However, this makes $n$ even so that necessarily $p=2$ and $n=12$.
