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ThudnBlunder
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Radicals  
« on: Apr 2nd, 2007, 1:49am »
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Find all positive numbers a such that 3(3 + a) + 3(3 - a) is an integer.
 
« Last Edit: Apr 2nd, 2007, 1:52am by ThudnBlunder » IP Logged

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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
« Last Edit: Apr 2nd, 2007, 7:43am by towr » IP Logged

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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?
 
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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

 
 
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« Last Edit: Apr 2nd, 2007, 1:29pm by SMQ » IP Logged

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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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Cheesy
 
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. Wink
 
 
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« Last Edit: Apr 4th, 2007, 10:59am by SMQ » IP Logged

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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].

What's the value of u3 - v3?
 
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