I saw this on Quora and have made very little progress.
What is the solution (exact or asymptotic) to $a_n =a_{n-1}^2+a_{n-2}^2, a_0, a_1>1 $?
I would be satisfied with a good asymptotic analysis.
Heck, I would be happy with an asymptotic form of the log of the solution.
A lower bound is easy.
Assuming $a_1 > a_0 > 1$,
$a_n>a_{n-1}^2 \gt a_{n-2}^4 ... > a_{n-k}^{2^k} ... \gt a_1^{2^{n-1}} \gt a_0^{2^{n}} $.
A reasonable upper bound seems harder.
$a_{n-1} > a_{n-2}^2 $ so $a_n =a_{n-1}^2+a_{n-2}^2 \lt a_{n-1}^2+a_{n-1} $ so $a_n+\frac14 \lt a_{n-1}^2+a_{n-1}+\frac14 =(a_{n-1}+\frac12)^2 $ or $a_n+\frac12 \lt (a_{n-1}+\frac12)^2+\frac14 $.
If $b_n = a_n+\frac12 $,
$\begin{array}\\ b_n &\lt b_{n-1}^2+\frac14\\ &\lt (b_{n-2}^2+\frac14)^2+\frac14\\ &= b_{n-2}^4+\frac12 b_{n-2}^2+\frac1{16}+\frac14\\ \end{array} $
Not sure where to go from this.
Another possibility is to notice that once $a_n$ gets large, for any $c > 0$ there is an $n(c)$ such that for $n \ge n(c)$, $a_n > 1/c$ so
$\begin{array}\\ a_n &=a_{n-1}^2+a_{n-2}^2\\ &\lt a_{n-1}^2+a_{n-1}\\ &= a_{n-1}^2(1+1/a_{n-1})\\ &\lt (1+c)a_{n-1}^2\\ &\lt (1+c)((1+c)a_{n-2}^2)^2\\ &=(1+c)^3a_{n-2}^4\\ &<(1+c)^3((1+c)a_{n-3}^2)^4\\ &=(1+c)^7a_{n-3}^8\\ & ...\\ &<(1+c)^{2^k-1}a_{n-k}^{2^k}\\ & ... \text{up to } n-k = n(c), k=n-n(c)\\ &<(1+c)^{2^{n-n(c)}-1}a_{n(c)}^{2^{n-n(c)}}\\ \end{array} $
Don't see how to do better.
For any $n \ge 1$, we have: \begin{align*} a_{n+1} &= a_n^2+a_{n-1}^2 \\ a_{n+1} &= a_n^2\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right) \\ \log a_{n+1} &= 2\log a_n + \log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right) \\ \dfrac{1}{2^{n+1}}\log a_{n+1} &= \dfrac{1}{2^n}\log a_n + \dfrac{1}{2^n}\log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right) \end{align*}
This is enough to show that $\dfrac{1}{2^n}\log a_n$ is non-decreasing, and thus, either converges to some limit or is unbounded.
Summing the previous equation from $n = 2$ to $n = N-1$ yields, $$\dfrac{1}{2^N}\log a_N = \dfrac{1}{4}\log a_2 + \sum_{n = 2}^{N-1}\dfrac{1}{2^n}\log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right).$$
It is easy to check that for all $n \ge 2$, we have $a_{n-1} > 1$, and thus, $a_n = a_{n-1}^2+a_{n-2}^2 > a_{n-1}^2 > a_{n-1}$. Hence, we can bound $$0 \le \sum_{n = 2}^{N-1}\dfrac{1}{2^n}\log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right) \le \sum_{n = 2}^{N-1}\dfrac{1}{2^n}\log 2 = \dfrac{1}{2}\log 2,$$ and thus, $$\dfrac{1}{4}\log a_2 \le \dfrac{1}{2^N}\log a_N \le \dfrac{1}{4}\log a_2 + \dfrac{1}{2}\log 2.$$
Since $\dfrac{1}{2^n}\log a_n$ is bounded (and previously shown to be non-decreasing), $\dfrac{1}{2^n}\log a_n$ converges to some number between $\dfrac{1}{4}\log a_2$ and $\dfrac{1}{4}\log a_2 + \dfrac{1}{2}\log 2$.
We can probably get tighter bounds if we are more careful about bounding the sum.