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Topic: Fun with algebraic number theory
Replies: 12   Last Post: Dec 7, 1999 7:08 PM

 Messages: [ Previous | Next ]
 Dave L. Renfro Posts: 575 Registered: 12/3/04
Re: Fun with algebraic number theory
Posted: Dec 4, 1999 3:22 PM

Dave Rusin <rusin@vesuvius.math.niu.edu>
[sci.math 3 Dec 1999 21:49:12 GMT]

wrote

>The following example was pointed out to me by John Wolfskill.
>
>(1) Show that sqrt( 8 + 3 sqrt(7) ) is the sum of the square
> roots of
> two integers.
>
>(2) Generalize.
>
>This came up as an illustration of some topics in Galois
>theory -- what looks like it ought to involve a dihedral
>Galois group lies in a field with group Z/2 x Z/2 -- but
>the topic expands a bit upon generalization, e.g., just
>which two quadratic extensions of Q have a compositum big
>enough to hold the required algebraic numbers.
>
>dave

You might want be interested in

Webster Wells, "Advanced Course in Algebra",
D. C. Heath and Co., 1904 [pp. 235-237]

<< BEGIN "QUOTE" ---->>>>>

ARTICLE 390: If sqrt[a + sqrt(b)] = sqrt(x) + sqrt(y),
where a, b, x, and y are rational expressions, and
a greater than sqrt(b), then
sqrt[a - sqrt(b)] = sqrt(x) - sqrt(y).

[proof omitted]

ARTICLE 391: The preceding principles may be used to find
the square root of certain expressions which are in the
form of the sum of a rational expression and a quadratic
surd.

Example: Find the square root of 13 - sqrt(160).

Assume, sqrt[13 - sqrt(160)] = sqrt(x) - sqrt(y) (1).

Then by #390, sqrt[13 + sqrt(160)] = sqrt(x) + sqrt(y).

Multiplying the equations gives sqrt(169 - 160) = x - y.
Hence, x - y = 3. Squaring (1), 13 - sqrt(160) =
x - 2*sqrt(xy) + y. Hence, x + y = 13. [Citing the
previously proved result that a + sqrt(b) = c + sqrt(d),
where a, c are rational and b, d are quadratic surds,
implies a=c and b=d.] It now easily follows that x=8
and y=5.

<<<<<<------ END "QUOTE" >>

Using this method for sqrt[8 + 3*sqrt(7)] gives

sqrt[8 + 3*sqrt(7)] = sqrt(9/2) + sqrt(7/2).

You can't write sqrt[8 + 3*sqrt(7)] as a sum
sqrt(x) + sqrt(y) with x and y both integers, however.
[There are not many pairs of integers to check. For
instance, among other restrictions, 1 <= x,y <= 8,
since the number in question is less than 4.]

Wells goes on to show that

sqrt[8 + sqrt(48)] = sqrt(6) + sqrt(2)

sqrt[22 - 3*sqrt(32)] = 3*sqrt(2) - 2

sqrt(392) + sqrt(360) = [2^(1/4)]*[3 + sqrt(5)]

ARTICLE 394: It may be proved, as in #390, that if
cube:root[a + sqrt(b)] = x + sqrt(y), where a, x are
rational expressions and sqrt(b), sqrt(y) quadratic
surds, then cube:root[a - sqrt(b)] = x - sqrt(y).

***************************************************
***************************************************

Some more exotic surd transformation techniques can
be found in:

Salvatore Composto, "Sulla trasformaxione del radicale
sqrt{a + sqrt[b + sqrt(c)]}", Peridico di Matematica
(3) 4 (1907), 32-36.

Salvatore Composto, "Sulla trasformazione dei radicali
sqrt[a +/- fourth:root(b)], sqrt[sqrt(a) +/- fourth:root(b)],
sqrt[fourth:root(a) +/- fourth:root(b)]", Peridico di
Matematica (3) 4 (1907), 75-83.

Here are some examples from these papers:

1. sqrt{12 + sqrt[38 + 2*sqrt(105)]}

= (1/2)*[sqrt(15) + sqrt(7) + sqrt(21) - sqrt(5)]

2. sqrt{3 + sqrt[-20 + 12*sqrt(5)]}

= (1/2)*[sqrt(10) + 2*fourth:root(5) - sqrt(2)]

3. sqrt{sqrt(2) + sqrt[-6 + 4*sqrt(3)]}

= (1/2)*[fourth:root(18) + fourth:root(24) - fourth:root(2)]

4. sqrt[7 + 2*fourth:root(132)]

= (1/2)*{sqrt[6 + 2*sqrt(33)] + sqrt[22 - 2*sqrt(33)]}

There are additional examples in the papers I
cited if anyone wants to pursue this further.

Dave L. Renfro

Date Subject Author
12/3/99 Dave Rusin
12/3/99 Kurt Foster
12/4/99 Robin Chapman
12/4/99 Pertti Lounesto
12/4/99 Denis Feldmann
12/4/99 Lee Rudolph
12/4/99 Pertti Lounesto
12/4/99 Virgil Hancher
12/5/99 Dave Rusin
12/7/99 Richard Bumby
12/4/99 Dave L. Renfro
12/5/99 Tapio Hurme
12/4/99 Dave L. Renfro