# Can the powers of 2 in the denominators of a fraction sum be used to find a contradiction? Let

Can the powers of 2 in the denominators of a fraction sum be used to find a contradiction?
Let ${a}_{1},{a}_{2},{b}_{1},{b}_{2},{c}_{1},{c}_{2},{d}_{1},{d}_{2},{e}_{1},{e}_{2}$ be odd integers, with
$gcd\left({a}_{1},{a}_{2}\right)=gcd\left({b}_{1},{b}_{2}\right)=gcd\left({c}_{1},{c}_{2}\right)=gcd\left({d}_{1},{d}_{2}\right)=gcd\left({e}_{1},{e}_{2}\right)=1,$
and assume $n\ge 3$ and $m\ge 1$ are integers such that
$\begin{array}{}\text{(}\star \text{)}& \frac{{a}_{1}}{{2}^{3\left(n-1\right)m}{a}_{2}}+\frac{{b}_{1}}{{2}^{3\left(\left(n-1\right)m+1\right)}{b}_{2}}+\frac{{c}_{1}}{{2}^{3\left(\left(n-1\right)m+1\right)}{c}_{2}}+\frac{{d}_{1}}{{2}^{3nm}{d}_{2}}+\frac{{e}_{1}}{{2}^{3nm}{e}_{2}}=1.\end{array}$
Is there a way to prove, using only the powers of 2 in the set of denominators, that (⋆) is impossible?
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Zichetti4b
The way it stands now, $\left(\star \right)$ is obviously possible. Let $n=3,m=2$. Then the denominators of five fractions contain (correspondingly, in the same order) . Now,
$\frac{184313}{45\cdot {2}^{12}}+\frac{1}{9\cdot {2}^{15}}+\frac{1}{1\cdot {2}^{15}}+\frac{1}{15\cdot {2}^{18}}+\frac{1}{1\cdot {2}^{18}}=\dots$
...guess what?
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