# A challenge for The Cult of Math

Fiathahel
Posers and Puzzles 03 Feb '04 11:10
1. Fiathahel
Artist in Drawing
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 ðŸ˜€ ).

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

Steffin
2. TheMaster37
Kupikupopo!
03 Feb '04 13:11
LOL

And if that's too easy, try the same for Q(sqrt( -10 + sqrt (10)) ).
3. Acolyte
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 ðŸ˜€ ).

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. Fiathahel
Artist in Drawing
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. Acolyte
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. Fiathahel
Artist in Drawing
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