4. Find all natural numbers $n$ whose set of proper divisors (i.e., all divisors except $n$) can be divided into two disjoint sets, each containing at least 2 elements, such that one set contains consecutive Fibonacci numbers, and the other set contains consecutive triangular numbers.

## First Grade - B Category
4. We will call the set of consecutive triangular numbers $T$, and the set of Fibonacci numbers $F$. Assuming that $1 \in T$. Then $3 \in T$. If $6 \in T$, it follows that $2 \in T$ (because then 2 is also a divisor of the number $n$, and its consecutive Fibonacci numbers, 1 and 3, are not in $F$), however, since 2 is not a triangular number, it follows that $6 \notin T$, thus we get $T=\{1,3\}$. Now, if we have two consecutive Fibonacci numbers $F_{k}$ and $F_{k+1}$ in $F$, then either $3 F_{k}$ or $3 F_{k+1}$ must also belong to the set $F$, because at least one of these two numbers is a proper divisor of the number $n$ (at most one of $F_{k}$, $F_{k+1}$ is divisible by 3, so if for example $3 \nmid F_{k}$, then $3 F_{k} \mid n$, but it cannot be $3 F_{k}=n$ because $F_{k+1} \mid n$ but $F_{k+1} \nmid 3 F_{k}$). However, for any Fibonacci number $F_{l}>1$, the number $3 F_{l}$ cannot be a Fibonacci number due to: $F_{l+3}=2 F_{l+1}+F_{l}>3 F_{l}>2 F_{l}+F_{l-1}=F_{l+2}$. It follows that $1 \notin T$.

Thus, $1 \in F$. Then $2 \in F$ must also be true. Suppose first $F=\{1,2\}$. Let $t$ be the smallest number in $T$. Then $t$ is a prime number, because otherwise some proper divisor of the number $t$ different from 1 and 2 (such a number definitely exists due to $t \neq 4$, since 4 is not a triangular number) would be a proper divisor of the number $n$, and would be either in $F$ or in $T$, but neither is possible. However, it is easy to see that 3 is the only triangular prime number. Thus, $3 \in T$, and then $6 \in T$. If $10 \in T$ as well, then 5 is neither a triangular number nor can it be in $F$ because $3 \notin F$, which is a contradiction. If only $T=\{3,6\}$, it follows that 6 is a proper divisor of the number $n$, so $n$ must be divisible by at least one more prime number different from 2 (since $4 \notin T, F$), but then there would exist a proper divisor of the number $n$ greater than 6, a contradiction.

Thus, we also have $3 \in F$. Then $6 \in T$ (because $6 \mid n$ and $n>6$, and 6 is not a Fibonacci number), and then $10 \in T$, so $5 \mid n$, and thus $15 \in T$ (a proper divisor, and not a Fibonacci number). If $21 \in T$ as well, then 7 would be a divisor of the observed number, but it is neither a Fibonacci number nor a triangular number. It follows that $T=\{6,10,15\}$, and from there $F=\{1,2,3,5\}$ (we cannot have $8 \in F$ because 4 is neither a Fibonacci number nor a triangular number) and finally $n=30$, which is the number that satisfies the conditions of the problem, and it is the only solution to the problem.

## First Grade - B Category