4. let the positive divisors of the natural number $n$ be $1=d_{1}<d_{2}<\ldots<d_{k}=n$. Determine all $n$ for which the following applies

$$
2 n=d_{5}^{2}+d_{6}^{2}-1
$$

## Solution
For each $1 \leq i \leq k$, $n / d_{i}$ is also a positive divisor of $n$ and a simple counting argument shows that this must be equal to $d_{k+1-i}$. Consequently, $n=d_{i} d_{k+1-i}$ applies for all $i$. The two estimates follow from the given equation

$$
\begin{gathered}
2 n=d_{5}^{2}+d_{6}^{2}-12 d_{5}^{2}
\end{gathered}
$$

This results in $d_{5}^{2}<n<d_{6}^{2}$, so according to the above remark, $n=d_{5} d_{6}$ must apply, in particular $k=10$. If this is inserted into the equation and transformed, then $1=\left(d_{6}-d_{5}\right)^{2}$ follows, i.e. $d_{6}=d_{5}+1$.

A natural number with the prime factorization $p_{1}^{\alpha_{1}} \cdots p_{r}^{\alpha_{r}}$ has exactly $\left(\alpha_{1}+\right.$ 1) $\cdots\left(\alpha_{r}+1\right)$ different positive divisors. Since $n$ has exactly $10=2 \cdot 5$ positive divisors
the prime factorization of $n$ must be of the form $n=p^{9}$ or $n=p^{4} q$ with different primes $p, q$. We now use that $d_{6}=d_{5}+1$ applies. Firstly, it follows that $d_{5}$ and $d_{6}$ are divisors, so $n$ is of the form $n=p^{4} q$. Secondly, exactly one of these divisors is even, so $q=2$ or $p=2$. In the first case, the divisor strangeness means that $d_{5}=2$ or $d_{6}=2$ applies. However, both are obviously not possible, since $d_{2}=2$ for sure. In the second case, $d_{5}=q$ or $d_{6}=q$ follows analogously. Because of $q \geq 3$, a brief inspection of the small divisors shows that $d_{6}=2^{4}$ or $d_{5}=2^{4}$. Only the second alternative leads to the prime number $d_{6}=17$ and to $n=2^{4} \cdot 17=272$. It is easy to calculate that this is indeed a solution to the problem.