[(a0000000…..000b)^2n - (b0000000…..000a)^2n] / (a^2 - b^2)is always divisible by _ and equal to __ ? {Where a>b, n is an even positive integer; a, b, n > 0} ?
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6sorry what is written here? – dato datuashvili Apr 10 '13 at 10:54
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CAPITAL LETTERS COMMENT. DO YOU ENJOY THIS? NEITHER DO WE ENJOY THE TITLE. – Asaf Karagila Apr 10 '13 at 10:56
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1@Asaf: I edited already. – Willie Wong Apr 10 '13 at 10:57
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@chilchin: What does it have to do with complex analysis? – Dennis Gulko Apr 10 '13 at 10:57
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u can assume l = k there or say there r 2n-2 zeroes in each term – chilchin Apr 10 '13 at 10:59
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ok no complex analysis for this question but see this one http://math.stackexchange.com/questions/356977/mathematical-problem-functional-analysis-no-approxiamation-is-required – chilchin Apr 10 '13 at 11:00
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What is the connection between those? – Dennis Gulko Apr 10 '13 at 11:01
1 Answers
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It's hard to tell exactly what OP wants, but note that ...
$$x^n-y^n = \left(x-y\right)\left(x^{n-1}+x^{n-2} y + \cdots + x y^{n-2} + y^{n-1}\right)$$
If we take
$$x := (a \cdot 10^m + b )^2 \qquad\qquad y := ( b \cdot 10^m + a )^2$$
and we consider just the first factor above, we have ...
$$\begin{align} x-y &=\left(a \cdot 10^m + b \right)^2 - \left( b \cdot 10^m + a \right)^2 \\ &= \left( a^2 \cdot 10^{2m} + 2 a b \cdot 10^m + b^2 \right) - \left(b^2 \cdot 10^{2m} + 2 a b \cdot 10^{m} + a^2 \right) \\ &= \left(a^2-b^2\right)\left(10^{2m}-1\right) \end{align}$$
Thus,
$$\frac{(a \cdot 10^m + b )^{2n} - ( b \cdot 10^m + a )^{2n}}{a^2-b^2}= (10^{2m}-1)\cdot\text{stuff} = \underbrace{999\dots9}_{2m \; \text{"$9$"s}} \cdot \text{stuff}$$
Blue
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good approach http://math.stackexchange.com/users/409/blue close enough , i am satisfied but i would rather be happy if you post the absolute final answer :) – chilchin Apr 10 '13 at 11:45
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1@chilchin: It's hard to post an absolute final answer when I'm not sure what the absolute initial question is. I've shown that the fraction is "always divisible by" $10^{2m}-1$; what else do you want? (The exact nature of "stuff" should be unimportant.) – Blue Apr 10 '13 at 11:49
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dude i said u can assume l = k there or say there r 2n-2 zeroes in each term there is a specific answer anyways http://math.stackexchange.com/users/409/blue i am leaving this site for bad governance njoy ur time here byee tc :) – chilchin Apr 10 '13 at 12:41
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1@chilchin: Dude, I used "$m$" in place of both "$k$" and "$l$", which amounts to assuming that those values are equal. :) – Blue Apr 10 '13 at 12:48
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yes right blue ...you will get a value in terms of a and b nevertheless its close and good try this question njoy bye http://math.stackexchange.com/questions/356977/mathematical-problem-functional-analysis-no-approxiamation-is-required – chilchin Apr 10 '13 at 12:54