any ideas?

Also Bulgaria IMO TST 2019 P1.

At first I thought this was an overkill, but on reflection, the main idea is quite intuitive: Firstly, rewrite the above as $a_{i+1}=2a_i^2-a_{i-1}$ and $b_{i+1}=2b_i^2-b_{i-1}$ for all $i=1,2,\dots,n-1$. Also note that $a_i,b_i\geq0$ for all $0\leq i\leq n$. Now set $c_i=2a_i$ and $d_i=2b_i$. Then $d_0=c_0$ and $c_{i+1}=c_i^2-c_{i-1}$ and $d_{i+1}=d_i^2-d_{i-1}$. Let $c=c_0=d_0$ and consider the following polynomials: \[f_1(x)=x, f_2(x)=x^2-c, f_3(x) = (x^2-c)^2-x, \dots\]\[f_{n+1}(x)=f_n(x)^2-f_{n-1}(x)\]Our aim is to prove that if $f_n(a_1)=f_n(b_1)$, then $a_1=b_1$. To do this, we will prove that for all $x$ for which $f_1(x),f_2(x),\dots,f_i(x)>=0$, the derivative of $f_i$ is positive. This would mean that $f_i$ is increasing in the interval considered and hence $f_k(c_1)=f_k(c_1)$, would imply $c_1=d_1$ and from there $a_1=b_1$. Before we do this, we will need the following 3 claims: Claim 1. $f_i(x)\geq 1$ for all $1\leq i\leq n-1$. Proof: For all $1\leq i\leq n-2$: \[f_{i-1}(x) = f_i(x)^2-f_{i+1}(x)\geq 0\]\[f_{i+2}(x) = f_{i+1}(x)^2-f_{i}(x)\geq 0\]so \[f_i(x)^2\geq f_{i+1}(x)\geq\sqrt{f_i(x)}\]so $f_i(x)\geq1$ for all $1\leq i\leq n-2$ Note that also $f_n(x)=f_{n-1}(x)^2 - f_{n-2}(x)\geq 0$ and so $f_{n-1}(x)^2 \geq f_{n-2}(x)\geq1$, so $f_{n-1}(x)\geq1$. Claim 2. $f_i(x)\geq 2$ for all $1\leq i\leq n-3$. Proof: For all $1\leq i\leq n-3$: \[f_{i-1}(x) = f_i(x)^2-f_{i+1}(x)\geq 1\]\[f_{i+2}(x) = f_{i+1}(x)^2-f_{i}(x)\geq 1\]so \[f_i(x)^2\geq 1+f_{i+1}(x)\geq1+\sqrt{1+f_i(x)}\]so $f_i(x)^2\geq 1+\sqrt{1+f_i(x)}$ which rewrites to $(f_i(x)-1)^2(f_i(x)-2)\geq0$, so $f_i(x)\geq2$ for all $1\leq i\leq n-3$. Claim 3. $2f_i'(x)\geq f_{i-1}'(x)$ for all $1\leq i\leq n-2$. Proof: We will prove this by induction on $i$/ For the base, clearly $f_1'(x)=1, f_2'(x)=2$ and $2f_3'(x)\geq f_2'(x)$. Now let $i\geq4$ and assume $2f_{i-1}'(x)\geq f_{i-2}'(x)$. We will prove that $2f_{i}'(x)\geq f_{i-1}'(x)$. For simplicity, let $f_{i-2}(x) = g, f_{i-1}(x)=f, f_i(x)=f^2-g, f_{i-3}(x) = g^2-f$. Hence by the induction hypothesis $f'\geq\frac{g'}{2}$ and since $f\geq2$, we have \[2ff' - \frac{f'}{2}\geq 4f'-\frac{f'}{2} = \frac{7f'}{2}\geq \frac{7}{4}g'\geq g'\]so $2ff' - \frac{f'}{2}\geq g'\Rightarrow (f^2-g)' = 2ff' - g'\geq \frac{f'}{2}$. Finally, let us finish the problem: Claim 4. $f_n(x)'>0$. Proof: For brevity, let $f_{n-1}(x) = f, f_{n-2}(x)=g, f_{n-3}(x) = g^2-f$. By the above claim, $2f_{n-2}'(x)\geq f_{n-3}'(x)$, so $2g'\geq 2gg'-f'$ and also $g^2-f\geq2$, so $g\geq\sqrt{2}$. Now we want $2ff'-g'\geq0$. Indeed, \[2ff' - g' \geq 2f'-g' \geq 2(2gg'-2g')-g' = g'(4g-5)\geq g'(4\sqrt{2}-5)>0\]With this the solution is complete. $\square$