Caleb Proctor

2022-07-15

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

$\begin{array}{}\text{(}\star \text{)}& \frac{{a}_{1}}{{2}^{3(n-1)m}{a}_{2}}+\frac{{b}_{1}}{{2}^{3((n-1)m+1)}{b}_{2}}+\frac{{c}_{1}}{{2}^{3((n-1)m+1)}{c}_{2}}+\frac{{d}_{1}}{{2}^{3nm}{d}_{2}}+\frac{{e}_{1}}{{2}^{3nm}{e}_{2}}=1.\end{array}$

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

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

$\begin{array}{}\text{(}\star \text{)}& \frac{{a}_{1}}{{2}^{3(n-1)m}{a}_{2}}+\frac{{b}_{1}}{{2}^{3((n-1)m+1)}{b}_{2}}+\frac{{c}_{1}}{{2}^{3((n-1)m+1)}{c}_{2}}+\frac{{d}_{1}}{{2}^{3nm}{d}_{2}}+\frac{{e}_{1}}{{2}^{3nm}{e}_{2}}=1.\end{array}$

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

Zichetti4b

Beginner2022-07-16Added 13 answers

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,

$\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?

$\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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