For $n=8$, start with a square of side length $4$, which clearly satisfies the problem's condition. Now we induct on n. For $n+1$, consider the figure of $n$ and just choose a side with length $\geq 2$ and "extend" the side by adding two new squares, that are adjacent to each other and no other side. Then the perimeter $p(n+1)$ is $p(n+1)=p(n)-2+4=p(n)+2=2(n+1)$ and the area $A(n+1)$ is $A(n+1)=A(n)+2=2(n+1)$. (For example, just the figure, that looks like a plus $+$ and extend one arm for each $n\longrightarrow n+1$)