# 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 $$\gcd(a_1,a_2)=\gcd(b_1,b_2)=\gcd(c_1,c_2)=\gcd(d_1,d_2)=\gcd(e_1,e_2)=1,$$ and assume $n \ge 3$ and $m \ge 1$ are integers such that $$\frac{a_1}{2^{3(n-1)m}a_2} + \frac{b_1}{2^{3\left((n-1)m+1\right)}b_2} + \frac{c_1}{2^{3\left((n-1)m+1\right)}c_2} + \frac{d_1}{2^{3nm}d_2} + \frac{e_1}{2^{3nm}e_2} = 1. \tag{\star}$$

Is there a way to prove, using only the powers of $2$ in the set of denominators, that ($\star$) is impossible?

• Oops! Corrected. Sep 22, 2016 at 14:57

The way it stands now, $(\star)$ is obviously possible. Let $n=3,m=2$. Then the denominators of five fractions contain (correspondingly, in the same order) $2^{12}, 2^{15}, 2^{15}, 2^{18},\text{ and }2^{18}$. Now, $${184313\over45\cdot2^{12}} + {1\over9\cdot2^{15}} + {1\over1\cdot2^{15}} + {1\over15\cdot2^{18}} + {1\over1\cdot2^{18}}=\dots$$ ...guess what?