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Topic: Radicals (Read 852 times) 

ThudnBlunder
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Find all positive numbers a such that ^{3}(3 + a) + ^{3}(3  a) is an integer.

« Last Edit: Apr 2^{nd}, 2007, 1:52am by ThudnBlunder » 
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Barukh
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Re: Radicals
« Reply #1 on: Apr 2^{nd}, 2007, 3:14am » 
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Should a be an integer?


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Grimbal
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Re: Radicals
« Reply #2 on: Apr 2^{nd}, 2007, 4:59am » 
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That would leave only the numbers 0 to 9 to try. I would say any number. "Positive" suggests a can not be complex. So any positive real number. There might be extra solutions if the result can be any Gaussian integer.


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towr
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Re: Radicals
« Reply #3 on: Apr 2^{nd}, 2007, 7:41am » 
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Isn't the answer all positive numbers? Like last time? For some integer x, and some y we have [x+ sqrt(y)]^{3}=3+sqrt(a) and [x  sqrt(y)]^{3}=3sqrt(a) x+ sqrt(y) + x  sqrt(y) = 2 x

« Last Edit: Apr 2^{nd}, 2007, 7:43am by towr » 
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Grimbal
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Re: Radicals
« Reply #4 on: Apr 2^{nd}, 2007, 10:03am » 
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I don't think you can always find such x and y. Consider a=0. ^{3}(3+a) + ^{3}(3a) = 2·^{3}3 and it is not an integer.


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SMQ
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Re: Radicals
« Reply #5 on: Apr 2^{nd}, 2007, 11:57am » 
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In fact, since a = 0 gives a value of 2^{3}3 2.88, and the function is monotonic decreasing and always positive, the only integer values it can have are 2 and 1 at the corresponding values of a, right? SMQ


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Barukh
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Re: Radicals
« Reply #6 on: Apr 2^{nd}, 2007, 12:06pm » 
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SMQ, I think you are right. The problem is to find an expression (I think in radicals) for such a.


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SMQ
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Re: Radicals
« Reply #7 on: Apr 2^{nd}, 2007, 12:24pm » 
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For x = ^{3}(3 + a) + ^{3}(3  a), I get a = 9  (x^{3}  6)^{3}/(3x)^{3}, which gives a = 242/27 at x = 2 and a = 368/27 at x = 1. Edit  derivation: (1) x = ^{3}(3 + a) + ^{3}(3  a) (2) x^{3} = 6 + 3^{3}((3 + a)^{2}(3  a) + 3^{3}((3 + a)(3  a)^{2}) (3) (x^{3}  6)/3 = ^{3}((3 + a)^{2}(3  a) + ^{3}((3 + a)(3  a)^{2}) (4) ((x^{3}  6)/3)^{3} = (3 + a)^{2}(3  a) + 3^{3}((3 + a)^{5}(3  a)^{4}) + 3^{3}((3 + a)^{4}(3  a)^{5}) + (3 + a)(3  a)^{2} (5) ((x^{3}  6)/3)^{3} = (3 + a)(3  a)[6 + 3^{3}((3 + a)^{2}(3  a) + 3^{3}((3 + a)(3  a)^{2})] (6) ((x^{3}  6)/3)^{3} = (9  a)[x^{3}] (substituting from (2) for the bracketed term) (7) 9  a = ((x^{3}  6)/(3x))^{3} SMQ

« Last Edit: Apr 2^{nd}, 2007, 1:29pm by SMQ » 
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Eigenray
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Re: Radicals
« Reply #8 on: Apr 2^{nd}, 2007, 4:31pm » 
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There's a simpler (or at least easier to read) derivation: If x = u + v, then x^{3} = u^{3} + v^{3} + 3uv x. Substituting then, we have x^{3} = 6 + 3(9a)^{1/3} x, which is easily solved for a.


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Barukh
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Re: Radicals
« Reply #9 on: Apr 4^{th}, 2007, 5:52am » 
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How about this variation: ^{3}(3 + a)  ^{3}(3  a) is an integer, and a is a rational?


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SMQ
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Re: Radicals
« Reply #10 on: Apr 4^{th}, 2007, 6:39am » 
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By Eigenray's derivation, x = u  v x^{3} = u^{3}  v^{3}  3uvx a = 9  (6  x^{3})/(3x)[/sup]3[/sup] so a is rational (and positive) for all integer x 0. Edit: erm, not quite, as Barukh points out below. See, this is why I'm a computer programmer and not a mathematician ... other programmers find my offthecuff answers convincing. SMQ

« Last Edit: Apr 4^{th}, 2007, 10:59am by SMQ » 
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Barukh
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Re: Radicals
« Reply #11 on: Apr 4^{th}, 2007, 10:49am » 
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on Apr 4^{th}, 2007, 6:39am, SMQ wrote:x = u  v x^{3} = u^{3}  v^{3}  3uvx[/hide]. 
 What's the value of u^{3}  v^{3}?


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