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

AI Thread Summary
The discussion revolves around finding integer values for a and b given that their least common multiple (lcm) is 2048. Initial calculations suggest a solution of a = 512 and b = -2048, but this leads to an inconsistency since 2 raised to any real number cannot be negative. Alternative solutions include a = 2048 and b = 1024, highlighting the symmetry in the problem. A systematic approach is proposed, involving the transformation of the equation into a quadratic form to derive integer solutions. The conversation emphasizes the need for careful manipulation of the equations to explore all potential integer pairs for a and b.
chwala
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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
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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.
 
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You have amde a mistake: if <br /> 2^{11 - y} = -1 then there is no solution for real y, since 2^x &gt; 0 for real x. Instead you have <br /> (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 <br /> \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 <br /> \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 <br /> 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 <br /> 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.
 
...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]##?
 
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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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