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We can prove that $x^2+x+1$ doesn't have any prime divisor of the form $3n-1$ and $x_{n+1}=1+x_0x_1...x_n$ so $x_{n+1} = 1 +A$ then $x_{n+2} = A^2+A+1$ and we are done because there are infinitely many $3n-1$ primes


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Suppose that $a_1 \cdots a_m \equiv 1 \pmod{p}$ for some prime $p$, which in particular implies that $p$ doesn't divide $a_1,\ldots,a_m$. We claim that $a_i \equiv a_{i \bmod m} \pmod{p}$ for $i > 0$, where the modulo returns an answer in the range $1,\ldots,m$. We prove this by induction. This is clear when $i \leq m$. Now suppose it holds for all $j <...


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Let me write it here, because I had a typo in the comment. First observe that $$ \begin{align} a_{n+2}&=1+a_0a_1\dotsm a_na_{n+1}\\ &=1+a_0a_1\dotsm a_n(1+a_0a_1\dotsm a_n) \end{align}$$ So $$4a_{n+2}=(2a_0a_1\dotsm a_n+1)^2+3$$ This implies that for an odd prime $p$ that divides $a_{n+2}$, we must have that $-3$ is a quadratic residue modulo $p$. By ...


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Hint: we have $A(x^2) = \sum_{n=0}^\infty a_nx^{2n} = a_0 + a_1x^2 + a_2x^4+ \cdots$. What can we say about the coefficients of odd powers ($x, x^3, \dots$)? Now do the same for $xB(x^2)$, what can we say about the its even power coefficients? What does this mean for their sum?


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