From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Fun with algebraic number theory
Date: 3 Dec 1999 21:49:12 GMT
Newsgroups: sci.math
Keywords: Some nested radical expressions simplify
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
==============================================================================
From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Fun with algebraic number theory
Date: 5 Dec 1999 20:08:57 GMT
Newsgroups: sci.math
Apology: I wrote
> The following example was pointed out to me by John Wolfskill.
...but the error was mine (I jumped too fast from the equation
sqrt(8+3 sqrt(7)) + sqrt(8-3 sqrt(7)) = 3 sqrt(2) .)
So let's turn this to
> (1) Show that sqrt( 8 + 3 sqrt(7) ) is the sum of the square roots of
two rational numbers.
In article <82b69o$trq$1@nnrp1.deja.com>,
Robin Chapman wrote:
>Do you really need algebraic nuumber theory to solve this sort of
>problem? Let's consider when sqrt(a + sqrt(b)) with a and b rational is
Of course there are other ways to do this, but this made for a nice example
for Wolfskill's Galois theory class. Let alpha = sqrt( epsilon ) where
epsilon = a + b sqrt(D) is an element of a quadratic extension of Q.
Clearly alpha satisfies the rational polynomial (X^2 - a)^2 - b^2 D = 0,
and except for some special cases this is its minimal polynomial. (For
example, we need epsilon itself not to be a square!) Now,
both alpha and -alpha lie in the field K = Q(alpha), of course, but
the other two roots beta = sqrt(a-b sqrt(D)) and -beta do not in
general lie in K -- in the generic case the normal closure of K is
a degree-8 extension of Q with dihedral Galois group. On the other hand,
alpha*beta = sqrt( Norm(epsilon) ) [the norm being from the quadratic
field down to Q], so in the special case that epsilon is a unit with
norm = 1, then alpha and beta (or -beta) are inverses, and in
particular, the minimal polynomial of alpha splits completely in K,
K/Q is Galois, and Gal(K/4) is a group of order 4. Clearly the
action of the Galois group on the set of these four roots is limited to
the obvious sign changes, from which we conclude Gal(K/Q) is the four-group
Z2 x Z2 rather than the cyclic group Z4.
But now there are three intermediate subgroups of index 2 and hence
K contains three quadratic extensions of Q, any two of which must then
have K as their compositum. OK, well it's clear that Q(sqrt(D)) is
one of the intermediate quadratic fields. But where are the other two?
They should be found by taking orbit-sums under the Galois group.
I note that alpha +- beta has a rational square 2(a+-1), so that
K contains the quadratic subfields sqrt(2(a+-1)). (You can get one from
the other by noting that if the three fields are Q(sqrt(d_i)) then
we must have d1 d2 = d3 * (a square); and (a+1) (a-1) = D * (b^2). )
When D is prime, there's not a lot of variation in the intermediate
quadratic fields, but for example taking epsilon = 6 + sqrt(35)
gives a little variation.
So yes, there are easier ways to finish the problem of writing some
of these numbers in the form sqrt(x)+sqrt(y). They're just not as
much fun :-)
dave
==============================================================================
From: bumby@lagrange.rutgers.edu (Richard Bumby)
Subject: Re: Fun with algebraic number theory
Date: 7 Dec 1999 19:08:48 -0500
Newsgroups: sci.math
In order to get by my news software, I need to shorten the quoted
material. Interested readers should track down the original postings.
Robin Chapman writes:
>In article <829dso$s46$1@gannett.math.niu.edu>,
> rusin@vesuvius.math.niu.edu (Dave Rusin) 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 --......
>Do you really need algebraic nuumber theory to solve this sort of
>problem? ...... We want to solve
>a + sqrt(b) = (sqrt(c) + sqrt(d))^2 ...... we have to have
>a = c + d, and sqrt(b) = 2 sqrt(c)sqrt(d) and so b/4 = cd. ......
>c and d are roots of a quadratic and can be found.
This approach to sqrt(a + sqrt(b)) goes back to Euclid, but it still
seems fresh. Modern notation makes it look easy, but it can appear in
unexpected places. In looking for some topics to lecture on while my
students were finishing the exercises assigned earlier, I was led to
reexamining something I published in volume 25 (1987, p. 62) of the
Fibonacci Quarterly, applying this to the "incredible identity" that
Dan Shanks published in vol. 12 (1974, p. 271) of the same journal.
Maybe you don't need algebraic number theory to solve the problem, but
there is a strong connection. Consider, for example, an article of
Shanks in the 7th SE conference on combinatorics, graph theory and
computing (1976).
--
R. T. Bumby ** Rutgers Math || Amer. Math. Monthly Problems Editor 1992--1996
bumby@math.rutgers.edu ||
Telephone: [USA] 732-445-0277 (full-time message line) FAX 732-445-5530
==============================================================================
From: dlrenfro@gateway.net (Dave L. Renfro)
Subject: Re: Fun with algebraic number theory
Date: 5 Dec 1999 01:41:23 -0500
Newsgroups: sci.math
Dave Rusin
[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
==============================================================================
From: John Wolfskill
Subject: answer to the sqrt 2 question
Date: Thu, 23 Dec 1999 15:43:25 -0600 (CST)
Newsgroups: [missing]
To: rusin
Dave,
At the West Coast NT Conference in Asilomar I presented in their problem
session the question I mentioned to you, about finding sqrt 2 in Q adjoin
sqrt \e, where \e is the fundamental unit (>1) of norm 1 in a real quadratic
numberfield Q adjoin sqrt d, with d \equiv 3 mod 4. Someone there pointed
out that having sqrt 2 in this field is equivalent to having a solution to
x^2 -d y^2 = \pm 2. Or, to put it another way, the prime lying over 2 is
principal iff sqrt 2 is in the field. Making the connection with a number
of norm \pm 2 is completely elementary; one writes sqrt \e as a rational
combination of sqrt 2 and sqrt (2d), which is what it would have to be,
and the coefficients of this combination lead, after trivial rescaling, to
a solution to the Pellian equation for \pm 2. This seems to me to be as
good an answer as one could expect to have. Determining which d have \pm 2
as a norm is likely to be very difficult, in analogy with the classically
deep and difficult question of when the fund unit has norm -1. Some others
suggested that one may be able to get density results, which again would
be in analogy with the norm -1 problem, and that sounds reasonable to me.
However, that wouldn't give more detailed information about what does or
does not happen with any specific value of d, and I think such information
simply is not possible to obtain.
Merry Christmas,
John