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Christine
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 Polynomial squaring ladders   « on: Nov 28th, 2013, 1:31pm » Quote Modify

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 29th, 2013, 12:16pm by Christine » IP Logged
towr
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 Re: Polynomial squaring ladders   « Reply #1 on: Nov 28th, 2013, 10:30pm » Quote Modify

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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Posts: 159
 Re: Polynomial squaring ladders   « Reply #2 on: Nov 29th, 2013, 12:57am » Quote Modify

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?

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 29th, 2013, 12:17pm by Christine » IP Logged
towr
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 Re: Polynomial squaring ladders   « Reply #3 on: Nov 29th, 2013, 4:58am » Quote Modify

on Nov 29th, 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 29th, 2013, 12:24pm » Quote Modify

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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 Re: Polynomial squaring ladders   « Reply #5 on: Nov 29th, 2013, 1:36pm » Quote Modify

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 » Quote Modify

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 7th, 2013, 12:48pm by Christine » IP Logged
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