1. A sequence of real numbers $a_{0}, a_{1}, a_{2}, \ldots$ satisfies the relation

$$
a_{m+n}+a_{m-n}-m+n-1=\frac{a_{2 m}+a_{2 n}}{2}
$$

for all $m, n \in \mathbb{N}, m \geqslant n$. If $a_{1}=3$, find $a_{2004}$.

(Kunap)
1. By setting $m=n$ we get $a_{0}=1$. Further, for $n=0$ we get $a_{2 m}=4 a_{m}-2 m-3$. Now, the equality from the problem becomes $a_{m+n}+a_{m-n}=2\left(a_{m}+a_{n}-n-1\right)$. From here, for $n=1$ we get $a_{m+1}-2 a_{m}+a_{m-1}=2$, i.e., $\quad b_{m}=b_{m-1}+2$ for all $m$, where $b_{m}=a_{m+1}-a_{m}$. Since $b_{0}=2$, it follows that $b_{m}=2 m+2$ and

$$
a_{m}=a_{0}+b_{0}+b_{1}+\cdots+b_{m-1}=1+(2+4+\cdots+2 m)=m^{2}+m+1
$$