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Topic: Polynomial squaring ladders (Read 3911 times) 

Christine
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Polynomial squaring ladders
« on: Nov 28^{th}, 2013, 1:31pm » 
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Step #1 : x^2  A^2 Step #2 : (x^2  A)^2  B^2 = (x^2  a^2)(x^2  b^2) Step #3 : ((x^2  A)^2  B)^2)  C^2 = (x^2  a^2)(x^2  b^2)(x^2  c^2)(x^2  d^2) Step #4 : (((x^2  A)^2  B)^2)  C^2)  D^2 = (x^2  a^2)...(x^2  h^2) If Step #1 : x^2  5^2 Step #2 : (x^2  17)^2  8^2 = (x^2  5^2)(x^2  3^2) what are Step #3 : ? Step #4 : ?

« Last Edit: Nov 29^{th}, 2013, 12:16pm by Christine » 
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towr
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Re: Polynomial squaring ladders
« Reply #1 on: Nov 28^{th}, 2013, 10:30pm » 
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Why is A 5 in the first step and 17 in the second? What are we trying to do here?


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Christine
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Re: Polynomial squaring ladders
« Reply #2 on: Nov 29^{th}, 2013, 12:57am » 
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on Nov 28^{th}, 2013, 10:30pm, towr wrote:Why is A 5 in the first step and 17 in the second? What are we trying to do here? 
 A = 5 and B = 7 work in steps 1 and 2. I couldn't somehow continue with these values onto steps 3 and 4. How? Can we continue with A=5, B=7, or maybe not? Is there a general formula to find A, B, C, D? A general formula that would allow us to add more steps?

« Last Edit: Nov 29^{th}, 2013, 12:17pm by Christine » 
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towr
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Re: Polynomial squaring ladders
« Reply #3 on: Nov 29^{th}, 2013, 4:58am » 
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on Nov 29^{th}, 2013, 12:57am, Christine wrote:A = 5 and B = 7 work in steps 1 and 2. 
 If I plug A = 5 and B = 7 into the formula you gave for step 2, that gives: (x^2  5)^2  7^2 = (x^2 + 2) (x^2  12) Hence why I don't understand what you're doing or trying to do.


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Christine
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Re: Polynomial squaring ladders
« Reply #4 on: Nov 29^{th}, 2013, 12:24pm » 
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Sorry towr, I made a mistake. Back to the question, how to come up with these types of polynomials: Step #1 : x^2  A^2 Step #2 : (x^2  A)^2  B^2 = (x^2  a^2)(x^2  b^2) Step #3 : ((x^2  A)^2  B)^2)  C^2 = (x^2  a^2)(x^2  b^2)(x^2  c^2)(x^2  d^2) Step #4 : (((x^2  A)^2  B)^2)  C^2)  D^2 = (x^2  a^2)...(x^2  h^2) On a personal note: I'm teaching myself math because I enjoy it. I'm trying the best I can. I'm holding two jobs, I don't have the time to attend college. Thanks for the feedback.


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towr
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Re: Polynomial squaring ladders
« Reply #5 on: Nov 29^{th}, 2013, 1:36pm » 
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For step two, we have (x^2  A)^2  B^2 = (x^2  A  B)(x^2  A + B) So we just need to find square AB and A+B, any two odd squares will do, e.g. 1+9 = 2A ==> A=5, 91=2B ==> B=4 So, (x^2  5)^2  4^2 = (x^2  3^2)(x^2  1^2) For step 3 we can simplify a bit, ((x^2  A)^2  B)^2)  C^2 = [(x^2  A)^2  B  C] * [(x^2  A)^2  B + C] But I don't see a way to move beyond that at the moment. Step 2 and 3 might not be possible at the same time, for all I know.


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Christine
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Re: Polynomial squaring ladders
« Reply #6 on: Dec 7^{th}, 2013, 12:31pm » 
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From step #2 to step #3 I cannot do it. But I found that to start step #3, all I needed was to find integers N that can be expressed as N = x^2  y^2 in 2 or more ways. Provided that the solutions (x,y) are of the same parity. The smallest integers that can be expressed in two ways are 9 and 15 5^2  4^2 = 3^2  0^2 = 9 8^2  7^2 = 4^2  1^2 = 15 It does not work for step #3 Step #3 : ((x^2  A)^2  B)^2)  C^2 = (x^2  a^2)(x^2  b^2)(x^2  c^2)(x^2  d^2) OR ((x^2  A)^2  B)^2)  C^2 = [(x^2  A)^2  B  C] * [(x^2  A)^2  B + C] But when N = 200 200 = x^2  y^2 x = ±51, y = ±49 x = ±27, y = ±23 x = ±15, y = ±5 solutions are all odd numbers. 15^2  5^2 = 51^2  49^2 15^2 + 49^2 = 51^2 + 5^2 = 2626 2626/2 = 1313 ((x^2  1313)^2  1421344)^2  237600^2 = (x^2  5^2)(x^2  15^2)(x^2  49^2)(x^2  51^2) A = 1313, B = 1421344, C = 237600 Note that we can also write: (x^2  a^2)^2  b^2 = (x^2  a^2  b)(x^2  a^2 + b) we just need to find square a^2  b and a^2 + b Right?

« Last Edit: Dec 7^{th}, 2013, 12:48pm by Christine » 
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