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Topic: Polynomial squaring ladders (Read 4130 times) |
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Christine
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Polynomial squaring ladders
« on: Nov 28th, 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 : ?
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| « Last Edit: Nov 29th, 2013, 12:16pm by Christine » |
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towr
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Re: Polynomial squaring ladders
« Reply #1 on: Nov 28th, 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 29th, 2013, 12:57am » |
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on Nov 28th, 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? |
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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?
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| « Last Edit: Nov 29th, 2013, 12:17pm by Christine » |
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towr
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Re: Polynomial squaring ladders
« Reply #3 on: Nov 29th, 2013, 4:58am » |
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on Nov 29th, 2013, 12:57am, Christine wrote:| A = 5 and B = 7 work in steps 1 and 2. |
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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 29th, 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 29th, 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 A-B and A+B, any two odd squares will do, e.g. 1+9 = 2A ==> A=5, 9-1=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 7th, 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?
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| « Last Edit: Dec 7th, 2013, 12:48pm by Christine » |
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