Sum of distinct members of a sequence lie in the same set

Question

Suppose a sequence is defined as $a_n=a_{n-2}-a_{n-1}$ with $a_0=-1$ and $a_1=2$. Now, do there exist any $m,n$ such that $a_m+a_n=a_{n'}$ except $m=n+1$ or $m=n-1$?, I.e., whether the sum of two non-consecutive terms of the sequence can be a term of the sequence? I think the difference equation satisfied by the sequence may give some clue, but am unable to get any simplification. Any hints. Thanks beforehand.

Answer 1

We can see that this sequence is the Fibonnaci sequence with every other term negated:

$$-1, 2, -3, 5, -8, 13, \ldots$$

So if two terms of this sequence added to another term of the sequence, we'd either have:

$$F_n + F_m = F_k$$ for three Fibonnaci numbers, or we'd have

$$F_n - F_m = F_k$$ for three Fibonnaci numbers, but then taking the $F_m$ across, we'd have again the first case. So, we just need to consider the equation:

$$F_n + F_m = F_k$$

WLOG, let's say $n > m$. Then we know that the only $m$ that allows the sum to be Fibonacci is $n-1$, else the sum falls too short. In other words, the only valid sum is:

$$F_n + F_{n-1} = F_{n+1}$$

To hammer the point home one more time, suppose $j < n - 1$. Then we have:

$$F_n < F_n + F_j < F_{n+1}$$

So $F_n + F_j$, being strictly between two consecutive Fibonnaci numbers, can't itself be part of the sequence.