Solving a recurrence relation with square root

Question

I ran into a bad recurrence relation. Anyone would calculate T(n) or add some hint?

$$T(n) = \begin{cases} n,\quad &\text{ if n=1 or n=0 }\\ \sqrt{1/2[T^2(n-1)+T^2(n-2)]+}n ,\quad &\text{ if n>1} \end{cases}$$

thanks to all

Answer 1

Hint: Consider the sequence $a(n)=T(n)^2$. This sequence satisfies the recurrence relation $a(0)=0, a(1)=1$ and $$ a(n) = T(n)^2=\frac{T^2(n-1)+T^2(n-2)}{2}+n = \frac{1}{2}a(n-1)+\frac{1}{2}a(n-2)+n, $$ for $n\geq 2$. Can you solve it from here?