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

• chwala
chwala
Gold Member
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.

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]##?

Last edited:
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 ##.

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