$m=n: a_{2m}+a_0=a_{2m} \to a_0=0$ $n=0: 4a_m=a_{2m}$ $a_{m+n}+a_{m-n}=2a_{m}+2a_n$ Let prove ,that $a_n=n^2$ by induction. $a_1=1,a_2=4,a_4=16,a_3=2a_2+2a_1-a_1=9$ $a_{n+1}=2a_n+2a_1-a_{n-1}=2n^2+2-n^2+2n-1=n^2+2n+1=(n+1)^2$ $a_{2007}=2007^2$

it is a really nice pre-olympiad problem congratulation

Easy::: Notice $a_n = n^2$ ===>>> $(m+n)^2+(m-n)^2=^? 1/2(4m^2+4n^2)$ Which is true hence $\boxed{a_{2007}=2007^2}$