Does it exist?

Algebra Level 3

Does there exist a sequence $\left\{ a_n \right\}_{n \in \mathbb{N}}$ that satisfies the following equation?

$a_i + a_j = a_{i-j}, \quad \forall i,j \in \mathbb{N}, i>j$

If it exists, then submit $a_{2018}$ , if it doesn't exist, submit $-1$ .

The answer is 0.

1 solution

Henry U
Dec 28, 2018

If we plug in $i=2n, j=n, n \in \mathbb{N}$ into the functional equation, we get

$\begin{aligned} a_{2n} + a_n & = a_{2n-n} \\ & = a_n \\ & \Rightarrow a_{2n} + a_n = a_n \\ & \Rightarrow a_{2n} = 0 \end{aligned}$

$2018 = 2 \cdot 1009 \Rightarrow a_{2018} = \boxed{0}$

Now, we plug in $i={n+2k}, j=n, k \in \mathbb{N}$

$a_{n+2k}+a_n=a_{2k}=0 \Leftrightarrow a_{n+2k}=-a_n$ .

Furthermore, we can compare $i=n+1, j=n$ to $i=n+2, j=n+1$

$a_{n+1}+a_n=a_1=a_{n+2}+a_{n+1} \Leftrightarrow a_{n+2} = a_n$

What we've now found out is

$a_{2n}=0$
$a_{n+2}=a_n$
$a_{n+2k}=-a_n$

From the second and third equation, for $k=1$ it follows

$a_{n+2}=a_n=-a_{n+2} \Leftrightarrow a_{n+2} = 0$

and therefore, the only sequence that satisfies all conditions is

$a_n = 0, \quad \forall n \in \mathbb N$

True but for the purposes of this problem, since you ask for the solution, don't you have to show that this is the only possible solution? Or at least that all possible solutions lead to $a_{2018}$ being 0?

Geoff Pilling - 2 years, 5 months ago

This is what I do in the first part of the solution. I only use given information (the functional equation) and derive $a_{2n} = 0$ , which is one part of the answer.

Then, I show that such a sequence actually exists by giving an example, tbe second part of the answer.

The question was "If such a sequence exists, then submit $a_{2018}$ , else submit $-1$ ." It does exist (part 2) and $a_{2018}$ is 0 (part 1).