Solving Radicals for A,B,C and D: 2^11-1

[ramsey2879]

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

(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).

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

Thanks for any pointers.

 

[ramsey2879]

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

(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).

I deduced this by studying the recursive sequence S^{n} = 3*S_{n-1} - 2*S_{n-2} 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 S_{n} = (C+1)*S_{n-1} - C*S_n always has embeded within two intermeshed series A_{i} = S_{2i} and B_{i} = S_{2i +1} which both are of the form S_{n} = (C^{2} + 1)S_{n-1} - C^{2}S_{n-2}. In the above series if we let S_0 = 0 then S_i|S_{ni}. I correctly deduced that this works recursively both forwards and backwords i.e if we let C^{'} = C^{2} etc. Thus there should is a recursive series where C = 2^{.25} and S_{44} = 2047S_4 Indeed If S_0 = 0 and S_1 = 1 then S_4 = (2^{.75}+2^{.50}+2^{.25} + 1) and S_44 = 2047S_4.

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

 

[Hurkyl]

The numbers 1, 2^{1/4}, 2^{1/2}, 2^{3/4} 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.

 

[ramsey2879]

The numbers 1, 2^{1/4}, 2^{1/2}, 2^{3/4} 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

 

[ramsey2879]

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

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

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

If we allow radicals of the form 2^{1/8} then S_3 can be factored as (1+2^{1/4} + 2^{1/8})*(1 + 2^{1/4} -2^{1/8})
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?

 

[DeaconJohn]

The numbers 1, 2^{1/4}, 2^{1/2}, 2^{3/4} 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