Find the possible values for ##a## and ##b## where ##a## and ##b## are integers

[chwala]

Homework Statement

Find possible values for $a$ and $b$ given that $a,b εℤ$

$\dfrac{1}{a}+\dfrac{1}{b}= \dfrac{3}{2048}$

Relevant Equations

Numbers

I noted that,

$lcm(a,b)=2048$


Letting $a=2^x$ and $b=2^y$,

$⇒\dfrac{1}{2^x}+\dfrac{1}{2^y}= \dfrac{3}{2^{11}}$

$⇒\dfrac{2^{11}}{2^x}+\dfrac{2^{11}}{2^y}= 3=[4-1]$

$⇒\dfrac{2^{11}}{2^x}+\dfrac{2^{11}}{2^y}=2^2-2^0$

My intention being to write all the numbers to base $2$.

$2^{11-x} + 2^{11-y} = 2^2-2^0$

$2^{11-x}=2^2, x=9$

$2^{11-y}=-2^0$
$-[2^{11-y}] = 2^0$
$11-y=0, y=11$

Therefore,

$a=2^9=512, b=-2^{11}=-2048$

Your insight is welcome...just picked up this question from internet.

Wondering if there are other possible values for $a$ and $b$...i need to check if there is a sequence of powers of $2$ for numerator part to make this a possibility.

 

[pasmith]

You have amde a mistake: if $$2^{11 - y} = -1$$ then there is no solution for real $y$, since $2^x > 0$ for real $x$. Instead you have $$(11 - y) \ln 2 = i\pi.$$ If you want to use this method, you must write $3 = 2^1 + 2^0$ not $3 = 2^2 - 2^0$.


We have $$\frac{3}{2048} = \frac{1}{2048} + \frac{2}{2048} = \frac{1}{2048} + \frac{1}{1024}$$ so that $a = 2048$, $b = 1024$ is a solution. Note also that the problem is symmetric, so if $(a,b)$ is a solution then so is $(b,a)$. Or we can do $$\frac{3}{2048} = \frac{4}{2048} - \frac{1}{2048} = \frac{1}{512} - \frac{1}{2048}.$$

A systematic approach is to multiply everything out to get $$2^{11}(a + b) = 3ab$$ so that $a + b = 3C$ and $ab = 2^{11}C$ for some integer $C$. Then $a$ and $b$ can be recovered as the roots of $$x^2 - 3Cx + 2^{11}C = \left( x - \frac{3C}{2}\right)^2 - \frac{9C^2 - 2^{13}C}{4} = 0.$$ We then require that $9C^2 - 2^{13}C = D^2$ is a perfect square, and for $a$ and $b$ to be integers we require $3C \pm D$ to be even.

 

[chwala]

...then can i manupilate my equation so that i now have,

$⇒\left[\dfrac{2^{11}}{2^x}+-\dfrac{2^{11}}{2^y}\right]=[4+-1]$?

$⇒[2^{11-x}+-2^{11-y} ]=[4+-1]$?

 

[Gavran]

The problem can be transformed into three problems.

The first one is $\frac {1}{a} = \frac {2^m}{2048}$ and $\frac {1}{b} = \frac {2^n}{2048}$ where $2^m+2^n = 3$ and $m, n \in \mathbb N_0$.

The second one is $\frac {1}{a} = \frac {2^m}{2048}$ and $\frac {1}{b} = -\frac {2^n}{2048}$ where $2^m-2^n = 3$ and $m, n \in \mathbb N_0$.

The third one is $\frac {1}{a} = -\frac {2^m}{2048}$ and $\frac {1}{b} = \frac {2^n}{2048}$ where $-2^m+2^n = 3$ and $m, n \in \mathbb N_0$.