$\prod_{i=1}^{n} (3 +\frac{1}{a_i})$ cannot be a power of 2 if $a_i \equiv \pm 1 \pmod 6$

Question

I have encountered this problem and I can't solve it.

If $a_i \equiv 1 \pmod 6$ or $a_i \equiv -1 \pmod \ 6$ and $a_i \ne \pm 1$, for every $i \in \{1, 2, .., n\}$, then prove that $$\prod_{i=1}^{n} \left(3 +\frac{1}{a_i}\right)$$ cannot be a power of $2$. Using induction, I proved that $$2^{n + 1} \lt \prod_{i=1}^{n} \left(3 +\frac{1}{a_i}\right)$$ Then I tried working$\mod 6$,$\mod4$ and $\mod 3$, but I couldn't solve it. Have you got any ideas?

Answer 1

This does not hold. Examples: $$(3+\frac1{-7})(3+\frac1{-5})=2^3$$ and $$(3+\frac1{17})(3+\frac1{19})(3+\frac1{29})(3+\frac1{143})(3+\frac1{215}) = 2^8.$$