$a_0=2a_0^2\to a_0=0$ or $a_0=\frac {1}{2}.$ 1) If $a_0=\frac{1}{2}$, then $a_i\equiv \frac 12$. $a_1=a_1^2+a_0^2\to (a_1-\frac{1}{2})^2=0\to a_1=\frac 12.$ $a_2=a_1^2+a_1^2=\frac 12$, $a_8=a_2^2+a_2^2=\frac{1}{2}$. By induction \[a_{2^{2^k-1}}=\frac 12.\] Because $a_n$ is increase $a_n\equiv \frac 12.$ $a_0=0$, then $a_1=0$ or $a_1=1$ 2) $a_0=0,a_1=0$, then $a_2=0\to a_8=0\to ...$, by motonocillaly $a_n\equiv 0.$ 3) $a_0=0, a_1=1$, then $a_2=2$, $a_4=4$, $a_5=5$, $a_8=8$,...$a_{2^{2j}}=a_{2^j}^2+0=2^j, a_{2^{2j+1}}=2a_{2^j}^2=2^{2j+1}.$ If $a_k=a$, then $a_{k^2}=a^2\to a_{k^2}=a_{k^2},...a_{k^{2^l}}=m^{2^l}=2^{2^l\log_2(m)}$. If $k>m$ we can find $l,j$, suth that $k^{2^l}>2^j, m^{2^l}<2^j$ - give contradition. If $k<m$ we can find $l,j$, suth that $k^{2^l}<2^j, m^{2^l}>2^j$ - contradition with increasing. Therefore $a_n\equiv n$.

hello Rust, what does this mean? Sonnhad.

nice solution Rust

Dr Sonnhard Graubner wrote: hello Rust, what does this mean? Sonnhad. hello, is your name Sonnhard or Sonnhad ? Graubner.

hello, my name is Sonnhard, it was a typo, sorry. Sonnhard.