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A challenge for The Cult of Math

A challenge for The Cult of Math

Posers and Puzzles

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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

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LOL

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

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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?

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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.

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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.

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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

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