
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. 235237]
<< 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), 3236.
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), 7583.
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

