Mathematical Reflections 2016, Issue 2 - Problem S368

Problem:
Determine all the natural numbers $n$ such that $$\sigma(n)=n+55,$$ where $\sigma(n)$ denotes the sum of the divisors of $n$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Since $\sigma(n) \geq n+1$, then the sum of the proper divisors of $n$ is $54$. Denoting by $\omega(n)$ the number of distinct primes dividing $n$, observe that if $\omega(n) \geq 4$, then there exist four prime numbers $p<q<r<s$ that divide $n$. Since $p \geq 2$, $q \geq 3$, $r \geq 5$ and $s \geq 7$, then $qrs$ divides $n$ but $qrs \geq 3\cdot5\cdot7>54$, which contradicts the hypothesis. So, $\omega(n) \leq 3$. We have three cases.

(i) $\omega(n)=1$. Then, $n=p^k$, where $p$ is a prime number and $k \in \mathbb{N}^*$. Hence, $$1+p+\ldots+p^k=p^k+55 \implies p(1+p+\ldots+p^{k-2})=54.$$ It follows that $p \in \{2,3\}$. An easy check shows that there are no solutions.

(ii) $\omega(n)=2$. Then, $n=p^{k_1}q^{k_2}$, where $p,q$ are prime numbers, $p<q$ and $k_1,k_2 \in \mathbb{N}^*$. Since $p \geq 2$ and $q \geq 3$, if $n=p^4q$, then $p+p^2+p^3+p^4+p^3q+q \geq 57>54$, so $k_1 \leq 3$. If $n=pq^3$, then $pq^2+q+q^2+q^3 \geq 2\cdot9+39>54$, so $k_2 \leq 2$.  We have six cases.

(a) $k_1=1,k_2=1$. Then, $n=pq$ and $p+q=54$, which gives $$(p,q) \in \{(7,47),(11,43),(13,41),(17,37),(23,31)\}.$$

(b) $k_2=2, k_2=1$. Then, $n=p^2q$ and $p+p^2+q+pq=54$. Both $p$ and $q$ must be odd. Since $p+p^2<54$, then $p \in \{3,5\}$. An easy check shows that there are no solutions.

(c) $k_1=3, k_2=1$. Then, $n=p^3q$ and $p+p^2+p^3+q+pq+p^2q=54$. Both $p$ and $q$ must be odd. Since $5^3>54$, then $p=3$. An easy check shows that there are no solutions.

(d) $k_1=1, k_2=2$. Then, $n=pq^2$ and $p+pq+q+q^2=54$. Since $q+q^2<54$, then $q \in \{3,5\}$. An easy check shows that there are no solutions.

(e) $k_1=2,k_2=2$. Then, $n=p^2q^2$ and $p+p^2+pq+p^2q+pq^2+q+q^2=54$. At least one between $p$ and $q$ must be even, so $p=2$. Thus, we get $$(p,q)=(2,3).$$

(f) $k_1=3, k_2=2$. Then, $n=p^3q^2$, but $p^3q+p^2q^2 \geq 8\cdot3+4\cdot9>54$, so there are no solutions.

(iii) $\omega(n)=3$. Then, $n=p^{k_1}q^{k_2}r^{k_3}$, where $p,q,r$ are prime numbers, $p<q<r$ and $k_1,k_2,k_3 \in \mathbb{N}^*$. Since $p \geq 2$, $q \geq 3$ and $r \geq 5$, if $n=p^3qr$, then $p^3q+p^3r \geq 8\cdot3+8\cdot5>54$, so $k_1 \leq 2$. If $n=pq^2r$, then $pq^2+q^2r \geq 2\cdot9+9\cdot5>54$, so $k_2=1$. If $n=pqr^2$, then $qr^2 \geq 3\cdot 25>54$, so $k_3=1$.  We have two cases.

(a) $k_1=k_2=k_3=1$. Then, $n=pqr$ and $p+q+r+pq+pr+qr=54$. Clearly, $p \neq 2$. If $p \geq 5$, then $q \geq 7$ and $r \geq 11$, but $pr=55>54$, contradiction. So, $p=3$ and $4q+4r+qr=51$, i.e. $(q+4)(r+4)=67$, but $67$ is a prime number, so no solutions in this case.

(b) $k_2=2, k_2=k_3=1$. Then, $n=p^2qr$ and $p+p^2+q+pq+p^2q+r+pr+p^2r+qr+pqr=54$. Clearly, $p$ cannot be even. So, $p \geq 3$ but $p^2r \geq 9\cdot7>54$, contradiction.


In conclusion, $n \in \{36,329,473,533,629,713\}$.