7. A set of positive integers is said to be pilak if it can be partitioned into 2 disjoint subsets $F$ and $T$, each with at least 2 elements, such that the elements of $F$ are consecutive Fibonacci numbers, and the elements of $T$ are consecutive triangular numbers. Find all positive integers $n$ such that the set containing all the positive divisors of $n$ except $n$ itself is pilak.
Solution.

The only positive integer $n$ that satisfies this property is 30 .
In this case, $D=\{1,2,3,5,6,10,15\}$, and we can partition $D$ into $F=\{1,2,3,5\}$ and $T=\{6,10,15\}$. We will show that there are no other $n$.

Claim 1: $1 \in F$.
Proof: Obviously, $1 \in D$. Assume that $1 \in T$. Since $T$ has at least 2 elements, $3 \in T$. If $6 \in T$, then $2 \in D$. Note that 2 is not a triangular number. If $2 \in F$, then since $F$ has at least 2 elements, at least one of 1 or 3 must be in $F$. But both 1 and 3 are in $F$, contradiction. Thus, $6 \notin T$, and $T=\{1,3\}$.
Now, if we have two consecutive Fibonacci numbers $F_{k}$ and $F_{k+1}$ in $F$, at least one of them is not divisible by 3 , so at least one of $3 F_{k}$ or $3 F_{k+1}$ divides $n$. It is well known that two consecutive Fibonacci numbers are relatively prime, so $F_{k+1} \mid 3 F_{k}$ and $F_{k} \mid 3 F_{k+1}$ are both impossible. Thus, $n \neq 3 F_{k}, 3 F_{k+1}$, implying that least one of $3 F_{k}$ or $3 F_{k+1}$ is in $F$. However, for all integers $i>1, F_{i+3}=F_{i+2}+F_{i+1}=2 F_{i+1}+F_{i}>3 F_{i}>2 F_{i}+F_{i-1}=$ $F_{i}+F_{i+1}=F_{i+2}$, so $3 F_{i}$ cannot be a Fibonacci number, and none of $3 F_{k}$ or $3 F_{k+1}$ is in $F$, contradiction. Therefore, $1 \in F$.

Claim 2: $3 \in F$.
If $1 \in F$, then since $F$ has at least two elements, $2 \in F$. If $3 \notin F$, then $F=\{1,2\}$. Let $t$ denote the smallest number in $T$. If $t$ is composite, since 4 is not triangular, there exists a proper divisor of $t$ greater than 2. Since this is a proper divisor of $n$, smaller than $t$, and greater than 2, this number cannot be in either $F$ or $T$, contradiction. Thus, $t$ is prime. However, 3 is the only triangular prime number, so $t=3$ and $3 \in T$. Since $1 \notin T$ and $T$ has at least two elements, $6 \in T$. If $10 \in T$, then $5 \in D$. Since 5 is not a triangular number, and $5 \notin F$, we have a contradiction. Thus, $10 \notin T$, and $T=\{3,6\}$. This implies that $D=\{1,2,3,6\}$, and that $n$ has exactly 5 divisors, making $n$ a perfect square. $n$ is even, thus it must be divisible by 4 . However, $4 \notin D$, contradiction. Therefore, $3 \in F$.

Since $1,3 \in F, 2 \in F$ and $6 \mid n$. Note that $n$ must have at least 5 positive integer divisors, and it is easy to check that among $1,2, \ldots, 15$, only 12 has at least 5 positive integer divisors. Thus, $n>6$ and $6 \in D$. Since 6 is not a Fibonacci number, $6 \in T$. Since $3 \notin T$ and $T$ has at least two elements, $10 \in T$. This also implies that $5 \in D$. Since 5 is not a triangular number, $5 \in F$. So $15 \mid n$. Since $5 \nmid 12, n \neq 12$, so $n>15$, making $15 \in D$. Since 15 is not a Fibonacci number, $15 \in T$. If $21 \in T$, then $7 \in D$. However, 7 is neither a Fibonacci number nor a triangular number, contradiction. Thus, $T=\{6,10,15\}$. If $8 \in F$, then $4 \in D$. However, 4 is neither a Fibonacci number nor a triangular number, contradiction. Thus, $F=\{1,2,3,5\}$.

Since $D=\{1,2,3,5,6,10,15\}, n=30$ is the only possible solution.