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Topic: Commutative Square Roots? (Read 380 times) |
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ThudnBlunder
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Commutative Square Roots?
« on: May 10th, 2004, 2:50am » |
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Find all integers a,b,c such that [smiley=surd.gif][a + (b/c)] = a[smiley=surd.gif](b/c)
Are there any non-integer solutions?
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| « Last Edit: May 10th, 2004, 3:23am by ThudnBlunder » |
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towr
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Re: Commutative Square Roots?
« Reply #1 on: May 10th, 2004, 4:26am » |
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on May 10th, 2004, 2:50am, THUDandBLUNDER wrote:| Are there any non-integer solutions? |
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Should a,b,c all be non-integers, or just any one of them? (because in the latter case it's rather trivial, take a'=a, b'=b/c and c'=1, where a,b,c are an integer solution)
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towr
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Re: Commutative Square Roots?
« Reply #2 on: May 10th, 2004, 4:37am » |
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::any c = b(a2 - 1)/a [ne] 0 seems to solve it for any a and b::
I suppose though that the 'commutative' in the title means there a '+' missing between a and [sqrt](b/c)
In which case the puzzle becomes somewhat harder.. 
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| « Last Edit: May 10th, 2004, 4:37am by towr » |
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Benoit_Mandelbrot
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Re: Commutative Square Roots?
« Reply #3 on: May 10th, 2004, 5:59am » |
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:: We can solve sqrt(x+a)=a*sqrt(x) for x, and we have x=a/(a^2-1). This means that b=a, and c=a^2-1 when a is an integer. There are infinitely many except when a=±1. a, b, and c are integers. There should be an infinite amount of non-integer solutions when a is irrational. When a is rational, being n/m, then b=n*m and c=n^2-m^2, when c is not equal to zero. Only b and c would be integers. ::
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| « Last Edit: May 10th, 2004, 6:02am by Benoit_Mandelbrot » |
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Sameer
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Re: Commutative Square Roots?
« Reply #4 on: May 10th, 2004, 12:49pm » |
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How is this commutative?
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ThudnBlunder
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Re: Commutative Square Roots?
« Reply #6 on: May 15th, 2004, 7:15pm » |
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It is the square root that is commuting! 
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