Cfrgtkky

Consider the following expression: $1/(sqrt a_1 ± sqrt a_2 ± ... sqrt a_n)$ where $a_1 ... a_n$ are positive integers

I would like a general algorithm by which I can convert expressions of this form into expression of the following form: exp/k

where k is some positive integer and exp is an expression consisting of additions, subtractions, multiplications and square roots but no divisions.

Here is an example of a simple case:

$1/(sqrt 2 + sqrt 3) = (sqrt 3 - sqrt 2)/(3 - 2) = sqrt 3 - sqrt 2$

Here I multiplied the top and bottom by the conjugate. This technique can be used in the general problem for n≤4 by breaking the sum up into pairs of roots. I will illustrate this below in the case where all square root terms are added rather than having subtraction mixed in.

consider this denominator: $(sqrt a + sqrt b) + (sqrt c + sqrt d)$

multiplying by its conjugate: $(sqrt a + sqrt b) - (sqrt c + sqrt d)$ yields:
$(sqrt a + sqrt b)^2 - (sqrt c + sqrt d)^2$
$a+b+2sqrt (ab) - c - d - 2sqrt (cd) = $
$ A + (sqrt B - sqrt C)$ where $A= a+b-c-d$ and $B=4ab$ and $C=4cd$

multiplying by its conjugate $ A - (sqrt B - sqrt C)$ yields:
$ A^2 - (sqrt B - sqrt C)^2 =$
$ A^2 - B - C + 2sqrt (BC) =$
$ D + sqrt E =$ where $D=A^2-B-C$ and $E=4BC$

now multiply by its conjugate $ D - sqrt E$ yields:
$ D^2 - E$ which is an integer

I would like to know how to generalize this to n=5 and beyond, or if this is not possible, why that is the case. Thanks

@ancientmathemetician:

I am not sure how to do this for the case n=8. Consider the following expression: $(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4) + (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)$

I could multiply by its conjugate which would yield the following

$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 - (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)^2$

However once each squared term is distributed out there will be 6 new square root terms introduced rather than just the original 4. Below I will do the first term:

$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 =$
$a_1 + a_2 + a_3 + a_4 + 2sqrt (a_1a_2) + 2sqrt (a_1a_3) + 2sqrt (a_1a_4) + 2sqrt (a_2a_3) + 2sqrt (a_2a_4) + 2sqrt (a_3a_4)$

as a result, I go from 8 roots in the denominator to 12 and a single non root term

If instead I broke up the expression into a group of 5 square roots and a group of 3 square roots I would end up with 10 roots from the group of 5 and 3 from the group of 3 totalling 13.

in either case the number of roots does not go down, which is the trick I have been exploiting. It seems to me a new trick of sorts is required to find the general solution.

Thanks

$$(sqrt a+sqrt b+sqrt c)(s+tsqrt a+usqrt b+vsqrt c+wsqrt{ab}+xsqrt{ac}+ysqrt{bc}+zsqrt{abc})=1$$
The rational coefficients of the various square-roots give you eight equations in eight unknowns $s$ to $z$.

It looks like it works for any number of square-roots. A $2^Ntimes2^N$ matrix has to be inverted. I don't know how to tell if it's going to be singular.

The matrix is very sparse - mostly zeros - so this isn't the most efficient method. I think it shows there is an answer.