Macedonian Mathematical Competitions

We have $a_n = \frac{n(n+1)}{2}$, $n \in \mathbb{N}$. Since $2a_m = a_n$, it follows that
$$2 \frac{m(m+1)}{2} = \frac{n(n+1)}{2}, \quad 2m(m+1) = n(n+1).$$
Now
$$\begin{align*} a_{2m-n} &= \frac{1}{2}(2m-n)(2m-n+1) = \frac{1}{2}(4m^2 - 4mn + n^2 + 2m - n) = \\ &= \frac{1}{2}\left(2m^2 + 2m + 2m^2 - 4mn + 2n^2 - n^2 - n\right) = \\ &= \frac{1}{2}(2m^2 - 4mn + 2n^2) = \frac{1}{2}2(m^2 - 2mn + n^2) = (m-n)^2 \end{align*}$$