Recurrence Relation(with Square root)
Let $f(n)=\sqrt{T(n)}$, $f$ satisfies the linear recurrence relation $f(n)=f(n-1)+2f(n-2)$. The  characteristic polynomial is $x^2-x-2$ and it's roots are $-1,2$, so $f(n)$ is a linear combination of $(-1)^n,2^n$. After substituting the initial conditions you get $f(n)=\frac{1}{3}(-1)^n+\frac{2}{3}2^n$.