A sequence $(a_n)_{n\ge0}$ satisfies $a_{m+n}+a_{m-n}=\frac12\left(a_{2m}+a_{2n}\right)$ for all integers $m,n$ with $m\ge n\ge0$. Given that $a_1=1$, find $a_{2003}$.
Problem
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Tags: algebra, functional equation
Wildabandon
10.04.2021 04:47
I think it also Russia 1995 problem
OlympusHero
10.04.2021 04:51
This is a Functional Equation in disguise. Now that I've said this, might well say it's an example in the Functional Equation chapter of IA.
kamatadu
23.05.2024 22:09
Here is a solution for this. Replace $m=n=0$ to get that $a_0 = 0$. Now we have $a_0 = 0$ and $a_1 = 1$. We proceed by using strong induction. Assume that $a_i = i^2$ for all $0\le i \le k$. We prove that $a_{k+1} = (k+1)^2$. Firstly, place $n = 0$ in the equation to derive that $a_{2m} = 4a_m$. Then our problem statement changes to $a_{m+n} + a_{m-n} = 2\left(a_m + a_n\right)$. Setting $m=k$ and $n=1$ gives $a_{k+1} + a_{k-1} = 2\left(a_k + a_1\right)$ which simplifies to $a_{k+1} = (k+1)^2$ and our induction is complete.