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Topic: Radicals (Read 852 times) |
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ThudnBlunder
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Find all positive numbers a such that 3(3 + a) + 3(3 - a) is an integer.
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« Last Edit: Apr 2nd, 2007, 1:52am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Barukh
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Re: Radicals
« Reply #1 on: Apr 2nd, 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 2nd, 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 2nd, 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=3-sqrt(a) x+ sqrt(y) + x - sqrt(y) = 2 x
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« Last Edit: Apr 2nd, 2007, 7:43am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Grimbal
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Re: Radicals
« Reply #4 on: Apr 2nd, 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(3-a) = 2·33 and it is not an integer.
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SMQ
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Re: Radicals
« Reply #5 on: Apr 2nd, 2007, 11:57am » |
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In fact, since a = 0 gives a value of 233 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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--SMQ
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Barukh
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Re: Radicals
« Reply #6 on: Apr 2nd, 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 2nd, 2007, 12:24pm » |
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For x = 3(3 + a) + 3(3 - a), I get a = 9 - (x3 - 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) x3 = 6 + 33((3 + a)2(3 - a) + 33((3 + a)(3 - a)2) (3) (x3 - 6)/3 = 3((3 + a)2(3 - a) + 3((3 + a)(3 - a)2) (4) ((x3 - 6)/3)3 = (3 + a)2(3 - a) + 33((3 + a)5(3 - a)4) + 33((3 + a)4(3 - a)5) + (3 + a)(3 - a)2 (5) ((x3 - 6)/3)3 = (3 + a)(3 - a)[6 + 33((3 + a)2(3 - a) + 33((3 + a)(3 - a)2)] (6) ((x3 - 6)/3)3 = (9 - a)[x3] (substituting from (2) for the bracketed term) (7) 9 - a = ((x3 - 6)/(3x))3 --SMQ
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« Last Edit: Apr 2nd, 2007, 1:29pm by SMQ » |
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Eigenray
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Re: Radicals
« Reply #8 on: Apr 2nd, 2007, 4:31pm » |
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There's a simpler (or at least easier to read) derivation: If x = u + v, then x3 = u3 + v3 + 3uv x. Substituting then, we have x3 = 6 + 3(9-a)1/3 x, which is easily solved for a.
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Barukh
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Re: Radicals
« Reply #9 on: Apr 4th, 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 4th, 2007, 6:39am » |
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By Eigenray's derivation, x = u - v x3 = u3 - v3 - 3uvx a = 9 - (6 - x3)/(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 off-the-cuff answers convincing. --SMQ
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« Last Edit: Apr 4th, 2007, 10:59am by SMQ » |
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Barukh
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Re: Radicals
« Reply #11 on: Apr 4th, 2007, 10:49am » |
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on Apr 4th, 2007, 6:39am, SMQ wrote:x = u - v x3 = u3 - v3 - 3uvx[/hide]. |
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