EMC 2020 Junior Category P2 wrote: A positive integer $k\geqslant 3$ is called fibby if there exists a positive integer $n$ and positive integers $d_1 < d_2 < \ldots < d_k$ with the following properties: $\bullet$ $d_{j+2}=d_{j+1}+d_j$ for every $j$ satisfying $1\leqslant j \leqslant k-2$, $\bullet$ $d_1, d_2, \ldots, d_k$ are divisors of $n$, $\bullet$ any other divisor of $n$ is either less than $d_1$ or greater than $d_k$. Find all fibby numbers. Proposed by Ivan Novak. It is trivial to check that $3$ and $4$ are fibby. For now on let us assume $k \geq 5$. We prove that in this case no fibby numbers exist. Suppose otherwise. Let $d_1<d_2 < \ldots <d_k$. We make the following Claim: Claim: All $d_i$ with $i \geq 3$ are odd. Proof: If there existed a $i \geq 3$ such that $d_i$ is even, then $\frac{d_i}{2}$ divides $n$ as well, hence $d_i<2d_1$ or $d_i>2d_k$. The latter is obviously impossible, while if the former holds then $$d_i<2d_1<d_1+d_2=d_3$$which contradicts the fact that $i \geq 3$ $\blacksquare$ To the problem, the Claim's result obviously poses a contradiction, since $d_5=d_3+d_4$ implies that at least one of the three $d_i$ involved must be even. Hence, we are done.

Suppose that $k\geq5$. Then we have that $x$, $y$, $x+y$, $x+2y$ and $2x+3y$ divide $n$, where $x$ and $y$ are positive integers, $y>x$. Consider the numbers $2(x+y)$ and $2y$. The first one is between $x+2y$ and $2x+3y$, and the second one is between $x+y$ and $x+2y$. Thus we have that $n=(x+y)k_1=yk_2$, where $k_1$ and $k_2$ are odd positive integers. Consider the $gcd(x, y)$, and let $x=dx_1$ and $y=dy_1$, where $x_1$ and $y_1$ are coprime. Case 1 $d$ is odd We have $(x_1+y_1)k_1=y_1k_2$. Then $y_1$ divides $k_1$, which is odd, i.e. $y_1$ is odd. Analogously, $x_1+y_1$ is odd. Then $x_1$ is even. But $n=d(x_1+y_1)k_1$ and all of these numbers are odd. Then $n$ is odd as well, but $x$ is even and divides it, absurd. Case 2 $d$ is even. Then $x$ and $y$ are even. Consider the number $(x+y)/2$. It is obviously between $x$ and $y$, and divides $n$, contradiction. Then we are left with $k=3$ and $k=4$. Now just take $n=30$, and the numbers $2,3,5$ and $1,2,3,5$, respectively.

We can check the case $k=3,4$ by hand. Now, we shall prove that all $k\geq 5$ is not a fibby number. Case 1: If $d_2-d_1$ is even then, let $d_2=d_1+2x$. Consider the first 3 terms of the sequence: \[d_1,d_1+2x,2d_1+2x\]meaning that $d_1+x$ divides $n$. However, $d_1<d_1+x<d_1+2x=d_2$. A contradiction due to the 3rd rule! $\blacksquare$ $\color{black} \rule{25cm}{2pt}$ Case 2: If $d_2-d_1$ is odd then, let $d_2=d_1+2x+1$. Subcase 1: If $d_1$ is even then, let $d_1=2y$. Consider the first 4 terms of the sequence \[2y,2y+2x+1,4y+2x+1,6y+4x+2\]meaning that $2(2y+2x+1)$ divides $n$. However, $d_3=4y+2x+1<2(2y+2x+1)<6y+4x+2=d_4$. A contradiction due to the 3rd rule! $\blacksquare$ $\color{black} \rule{25cm}{2pt}$ Subcase 2: If $d_1$ is odd then, let $d_1=2y+1$. Consider the first 3 terms of the sequence \[2y+1,2y+2x+2,4y+2x+3\]meaning that $2(2y+1)$ divides $n$. However, $d_1=2y+1<2(2y+1)<4y+2x+3=d_3$. By the 3rd rule we have, $d_2=2(2y+1)$ which is equivalent to $x=y$. Now, let's see the first 5 terms of the sequence: \[2x+1,4x+2,6x+3,10x+5,16x+8\]This means that $8x+4$ divides $n$. However, $d_3=6x+3<8x+4<10x+5=d_4$. A contradiction due to the 3rd rule! $\blacksquare$

Orestis_Lignos wrote: EMC 2020 Junior Category P2 wrote: A positive integer $k\geqslant 3$ is called fibby if there exists a positive integer $n$ and positive integers $d_1 < d_2 < \ldots < d_k$ with the following properties: $\bullet$ $d_{j+2}=d_{j+1}+d_j$ for every $j$ satisfying $1\leqslant j \leqslant k-2$, $\bullet$ $d_1, d_2, \ldots, d_k$ are divisors of $n$, $\bullet$ any other divisor of $n$ is either less than $d_1$ or greater than $d_k$. Find all fibby numbers. Proposed by Ivan Novak. It is trivial to check that $3$ and $4$ are fibby. For now on let us assume $k \geq 5$. We prove that in this case no fibby numbers exist. Suppose otherwise. Let $d_1<d_2 < \ldots <d_k$. We make the following Claim: Claim: All $d_i$ with $i \geq 3$ are odd. Proof: If there existed a $i \geq 3$ such that $d_i$ is even, then $\frac{d_i}{2}$ divides $n$ as well, hence $d_i<2d_1$ or $d_i>2d_k$. The latter is obviously impossible, while if the former holds then $$d_i<2d_1<d_1+d_2=d_3$$which contradicts the fact that $i \geq 3$ $\blacksquare$ To the problem, the Claim's result obviously poses a contradiction, since $d_5=d_3+d_4$ implies that at least one of the three $d_i$ involved must be even. Hence, we are done. In your claim , di/2 can be any dj ( 1=< j<=k) .