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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?

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  • $\begingroup$ What if $a_i=-1$ for all $i$? If the $a_i$ have to be positive, then the product is at least $3^n$, so if your proof by induction was correct, it follows that $3^n\le 2^{n+1}$ and $(3/2)^n\le 2$ which means that $n=1$. $\endgroup$
    – Mastrem
    Jun 20 '21 at 14:30
  • $\begingroup$ Oh, well, I forgot about $-1$. $\endgroup$
    – andu eu
    Jun 20 '21 at 14:34
  • $\begingroup$ My mistake, I have revised the proof and it contained a mistake. $\endgroup$
    – andu eu
    Jun 20 '21 at 14:59
  • $\begingroup$ So is it true and given that this product is indeed bounded by $2^{n+1}$? OP, it is on the person writing the question to be clear, and this is hard to follow with the corrections made in the comments. $\endgroup$
    – Mike
    Jun 20 '21 at 15:18
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    $\begingroup$ Well, first of all, the case for $n = 1$, $3 + \frac{1}{a_1}$ is clearly not a power of two. If $n = 2$, $(3 + \frac{1}{a_1})(3 + \frac{a_2}) > 3^2 > 2^3$. Now assume $\prod_{i = 1}^{n}(3 + \frac{1}{a_i}) > 2^{n + 1}$ and show that $\prod_{i = 1}^{n + 1}(3 + \frac{1}{a_i}) > 2^{n + 2}$. But $\prod_{i = 1}^{n + 1}(3 + \frac{1}{a_i}) > 2^{n + 1} (3 + \frac{1}{a_{n+1}}) > 2^{n + 2}$. The other assumptions I made were false. $\endgroup$
    – andu eu
    Jun 20 '21 at 15:42
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This does not hold. Examples: $$(3+\frac1{-7})(3+\frac1{-5})=2^3$$ and $$(3+\frac1{17})(3+\frac1{19})(3+\frac1{29})(3+\frac1{143})(3+\frac1{215}) = 2^8.$$

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