 # A challenge for The Cult of Math Fiathahel Posers and Puzzles 03 Feb '04 11:10
1. 03 Feb '04 11:10
Find all elements in Q(sqrt(5)) which are a root of a polynomial with integer coefficient (and post them in this thread 😀 ).

Q(sqrt(5)) is the smallest field containing the rationals and sqrt(5).

Steffin
2. 03 Feb '04 13:11
LOL

And if that's too easy, try the same for Q(sqrt( -10 + sqrt (10)) ).
3. 06 Feb '04 11:42
Originally posted by Fiathahel
Find all elements in Q(sqrt(5)) which are a root of a polynomial with integer coefficient (and post them in this thread 😀 ).

Q(sqrt(5)) is the smallest field containing the rationals and sqrt(5).

Steffin
All of them are algebraic. What are you getting at?
4. 06 Feb '04 12:12
Originally posted by Acolyte
All of them are algebraic. What are you getting at?
Of course they are algebraic. But the task is to find them.
5. 06 Feb '04 15:56
Originally posted by Fiathahel
Of course they are algebraic. But the task is to find them.
Q(sqrt(5)) is all numbers of the form a + b*sqrt(5), where a and b are rationals. Do you mean produce polynomials that have these numbers as roots? Well, p/q + (r/s)*sqrt(5) is a root of the following:

f(x) = QSX - 2pqSx + (PS - 5RQ) = 0

where P, Q etc mean p^2, q^2 etc.
6. 06 Feb '04 16:341 edit
Originally posted by Acolyte
Q(sqrt(5)) is all numbers of the form a + b*sqrt(5), where a and b are rationals. Do you mean produce polynomials that have these numbers as roots? Well, p/q + (r/s)*sqrt(5) is a root of the following:

f(x) = QSX - 2pqSx + (PS - 5RQ) = 0

where P, Q etc mean p^2, q^2 etc.
Sorry Acolyte, I forgot something. You were right all of them met the criteria. I forgot to say that the leading coefficient of the polynomial had to be 1.
So again:

Find all elements in Q(sqrt(5)) which are a root of a polynomial with integer coefficient and leading coefficient 1