Fibonacci Sums

Solution 1: Direct considerations

After some trial and error, we suspect that the answer is "no solutions". How can we show this?
We have the following observations:

1. There are always real solutions to $a + b = F_x, b + c = F_y , c + a = F_z$ , namely $a = \frac{ F_x + F_z - F_y } { 2 } , b = \frac{ F_x + F_y - F_z } { 2} , c = \frac{ F_y + F_z - F_x } { 2 }$ . Hence, we have to investigate the additional conditions of positive and integer .
2. For integer solutions, we just have to ensure that $F_x + F_y + F_z$ is even, which is easily satisfied. This is unlikely to yield anything as yet.
3. Thus, if there are no solutions, it is because one of them is NOT positive. WLOG, let $a < b < c$ . Then, we want to show that $0 \geq a$ , or equivalently that $0 \geq F_x + F_z - F_ y$ .

Since we have $F_y > F_z > F_x$ , so $z \leq y -1$ and $x \leq y -2$ . Thus,

$F_x + F_z - F_y \leq F_{y-2} + F_{y-1} - F_y = 0$

as desired. In all solutions to the system, $a$ must be non-positive, thus there are no distinct positive integer solutions.

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Solution 2: Proof by contradiciton

Suppose that there are positive integer solutions to $a + b = F_x, b + c = F_y , c + a = F_z$ .
WLOG, $1 \leq a < b < c$ .
Since $F_y = c + b > c + a = F_z$ , so $y -1 \geq z$ .
We have $b - a = F_y - F_z \geq F_y - F_{y-1} = F_{y-2}$ .
So, $F_{y-2} \leq b - a < b + a < a + c < b+c = F_y$ .

We have 2 Fibonacci numbers between $F_{y-2}$ and $F_y$ , which is a contradiction.

Note: This solution arose from the realization that $c+a$ and $c + b$ being Fibonacci numebrs puts a restriction on the size of $b$ .

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Note: If the condition of "positive" is relaxed to allow for 0, then equality must hold throughout and thus the only solution is the obvious solution $\{ 0, F_n, F_{n+1} \}$ .