Radicals and 2^11-1

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I am weak in this topic but I am certain that the following is has a solution in integers A,B,C,D.

Find A,B,C and D as integers such that

[tex] (3*2^{.75} + 7*2^{.50} + 7*2^{.25} + 7)*(A*2^{.75} + B*2^{.50} + C*2^{.25} + D) = 2047*(2^{.75} + 2^{.50} +2^{.25} + 1) [/tex].

I deduced this by studying the recursive sequence [tex]S^{n} = 3*S_{n-1} - 2*S_{n-2}[/tex] That is {1,3,7,15,31, ...}

Thanks for any pointers.
 
841
0
I am weak in this topic but I am certain that the following is has a solution in integers A,B,C,D.

Find A,B,C and D as integers such that

[tex] (3*2^{.75} + 7*2^{.50} + 7*2^{.25} + 7)*(A*2^{.75} + B*2^{.50} + C*2^{.25} + D) = 2047*(2^{.75} + 2^{.50} +2^{.25} + 1) [/tex].

I deduced this by studying the recursive sequence [tex]S^{n} = 3*S_{n-1} - 2*S_{n-2}[/tex] That is {1,3,7,15,31, ...}

Thanks for any pointers.
Sorry but I couldn't do the math associated with the above problem. I will give an explanation of how I came up with it though. The recursive series [tex]S_{n} = (C+1)*S_{n-1} - C*S_n[/tex] always has embeded within two intermeshed series [tex]A_{i} = S_{2i}[/tex] and [tex]B_{i} = S_{2i +1}[/tex] which both are of the form [tex]S_{n} = (C^{2} + 1)S_{n-1} - C^{2}S_{n-2}[/tex]. In the above series if we let [tex]S_0 = 0[/tex] then [tex]S_i|S_{ni}[/tex]. I correctly deduced that this works recursively both forwards and backwords i.e if we let [tex]C^{'} = C^{2}[/tex] etc. Thus there should is a recursive series where [tex]C = 2^{.25}[/tex] and [tex]S_{44} = 2047S_4[/tex] Indeed If S_0 = 0 and S_1 = 1 then [tex] S_4 = (2^{.75}+2^{.50}+2^{.25} + 1)[/tex] and [tex]S_44 = 2047S_4[/tex].

[tex]S_{11} = 3*2^{.75} + 7*2^{.50} + 7^2^{.25} + 7[/tex] and of course S_11 divides S_44!
 

Hurkyl

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The numbers [itex]1, 2^{1/4}, 2^{1/2}, 2^{3/4}[/itex] are linearly independent over Q. Therefore, you can split your equation into four equations in four unknowns, and so you would usually expect to be able to apply linear algebra to compute the unique (rational) solution.
 
841
0
The numbers [itex]1, 2^{1/4}, 2^{1/2}, 2^{3/4}[/itex] are linearly independent over Q. Therefore, you can split your equation into four equations in four unknowns, and so you would usually expect to be able to apply linear algebra to compute the unique (rational) solution.
thanks

solution A= 4, B = 32, C = 256, D = 1
 
841
0
Another question,
My series is of the form [tex]S_{n} = G*S_{n-1} + H*S_{n-2}[/tex] where [tex]S_0}=0[/tex] and [tex]S_{1}=1[/tex].
Also [tex]S_{n} = \{A,B,C,D\} = A + B*2^{1/4}+C2^{1/2} +D2^{3/4}[/tex]
If a = floor n/4 then S_n
=
[tex] \{2^{a}-1,2^{a}-1,2^{a}-1,2^{a}-1\} [/tex], or
[tex] \{2^{a+1}-1,2^{a}-1,2^{a}-1,2^{a}-1\}[/tex], or
[tex]\{2^{a+1}-1,2^{a+1}-1,2^{a}-1,2^{a}-1\}[/tex], or
[tex]\{2^{a+1}-1,2^{a+1}-1,2^{a+1}-1,2^{a}-1\}[/tex], depending upon
the value of n mod 4, as 0,1,2,3 respectively.

If [tex] p = \prime > 3[/tex] then S_3 divides [tex]S_{4p+1}[/tex] or [tex]S_{4p+2}[/tex]

So If p = -1 mod 4 both S_2 and S_3 divide S_(4p+2).

If we allow radicals of the form [tex]2^{1/8}[/tex] then [tex]S_3[/tex] can be factored as [tex](1+2^{1/4} + 2^{1/8})*(1 + 2^{1/4} -2^{1/8})[/tex]
Also S_2 can be factored as (1+2^{1/4})*(1+2^{1/2} and each factor can be also be expressed as the difference of two squares if we allow units in "i".
And of course as I showed earlier S_(2p) divides S_(4p) and S_2 divides both S_2p and S_4p.
Could any of this be useful in factoring 2^{p} -1?
 
Last edited:
122
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The numbers [itex]1, 2^{1/4}, 2^{1/2}, 2^{3/4}[/itex] are linearly independent over Q. Therefore, you can split your equation into four equations in four unknowns, and so you would usually expect to be able to apply linear algebra to compute the unique (rational) solution.

Hurkyl, Really nice! DJ
 

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