A toy question on solutions to a sequence of trinomials over $\mathbb{F}_3$ Suppose that $\{a_{i}\}_{i\ge1}$ and $\{b_{i}\}_{i\ge1}$ are sequences
of natural numbers such that for each $x\in\overline{\mathbb{F}_{3}}^{*}$
there is a natural number $I_{x}$ satisfying that $x^{a_{i}}+x^{b_{i}}+1=0$
for each $i\ge I_{x}$. Does it follows that for each $x\in\overline{\mathbb{F}_{3}}^{*}$
there is a natural number $I'_{x}$ satisfying that $x^{a_{i}}=x^{b_{i}}=1$
for each $i\ge I'_{x}$?
Behind this question is the assertion that vaguely says there is only one solution, i.e. (1,1), to the equation X+Y+1=0 in the adelic closure of the cyclic subgroup of $\mathbb{F}_{3}(T)^*$ generated by $T$.
(Note that this assertion with the words "the adelic closure of" deleted is obviously true.)
 A: Equivalently, for every $n$ there exists $I_n$ such that $x^{a_i}+x^{b_i}+1$ vanishes on $GF(3^n)^*$ for all $i\geq I_n$, which means that $x^{a_i}+x^{b_i}+1$ is divisible by $x^{3^n-1}-1$. On the other hand, the question is whether $3^n-1$ divides $a_i$ and $b_i$ in this case. The answer is clearly yes, otherwise there would be a residue class modulo $3^n-1$ containing exactly one of $a_i,b_i,1$, which contradicts the given divisibility of polynomials.
A: For any natural $k$, non divisible by 3, there exists $n$ such that each root of $x^{k}-1$ is a root of $x^{a_i}+x^{b_i}+1$. Since $f_k(x)=x^k-1$ does not have double roots (its derivative $f_k'=kx^{k-1}$ is coprime to $f_k$), we conclude that $f_k$ divides $x^{a_i}+x^{b_i}+1$. Reduction of $x^a$ modulo $x^k-1$ is equivalent to reduction of $a$ modulo $k$. Therefore $x^{\alpha_i}+x^{\beta_i}+1$ should be divisible by $f_k$, where $\alpha_i,\beta_i$ are remainders of $a_i$, $b_i$ modulo $k$. This polynomial has degree strictly less than $k$, hence it must be equal to 0. This is possible only if $\alpha_i=\beta_i=0$, else the value at 0 is non-zero. Therefore we have proven that $k$ divides all $a_i,b_i$ for large enough $i$. But any $x\ne 0$ from algebraic closure of $\mathbb{F}_3$ satisfies $x^k=1$ for some $k$ non-divisible by 3. Thus your statement.
Now assume that all but finitely many $x\in \overline{\mathbb{F}_{3}}^{*}$ satisfy $x^{a_i}+x^{b_i}+1$ for large $i$. All these finitely many $x$ satisfy the same equation $x^A=1$ for some $A$ coprime to 3. Choose large $B$ coprime to 3 and consider the following polynomial $g(x)=(x^{AB}-1)/(x^A-1)=1+x^A+\dots+x^{A(B-1)}$. We have $g(x)\equiv B$ modulo $x^A-1$, hence $g(x)$ and $x^A-1$ are coprime. That is, for large $i$ polynomial $g(x)$ divides $x^{a_i}+x^{b_i}+1$. We may reduce $a_i,b_i$ modulo $AB$. Denote remainders $\alpha_i,\beta_i$. Write down $x^{\alpha_i}+x^{\beta_i}+1=g(x)h(x)=(1+x^A+\dots+x^{A(B-1)})(c_0+c_1x+\dots+c_{A-1}x^{A-1})$. Expanding brackets on the right hand side we see that nothing cancels. Thus it may be possible only if $h\equiv 0$ and $\alpha_i=\beta_i=0$, i.e. $a_i,b_i$ are divisible by $AB$. Since $B$ is arbitrary number coprime to 3, it implies that any $x\in \overline{\mathbb{F}_{3}}^{*}$ satisfies $x^{a_i}+x^{b_i}+1$ for large $i$. 
