Algebra Precalculus – Polynomial with Rational Coefficients and Root sqrt(2)+sqrt(3)-sqrt(5)

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Form a polynomial of smallest degree having rational coefficients and one root as $\sqrt{2}+\sqrt{3}-\sqrt{5}$

Idea 1:

I thought that other roots would be just different combination of signs on the surds, ie

  • $\sqrt{2}+\sqrt{3}+\sqrt{5}$
  • $\sqrt{2}-\sqrt{3}+\sqrt{5}$

so least degree would be $2^3 = 8$.

Polynomial then could be formed using viete's formulas.

Idea 2:

We let $x = \sqrt{2}+\sqrt{3}-\sqrt{5}$. Then rearranging and squaring repeatedly gives us the polynomial.

Questions

  1. This method seems unsatisfactory and is just a thought. Please help me with a proper method.

  2. Is the polynomial i found unique? or there are more polynomials with rational coefficients with this root ($\sqrt{2}+\sqrt{3}-\sqrt{5}$)?

  3. Also can we generalise this result: that the least degree of a polynomial whose root is a sum of $n$ distinct surds is $\sum \binom{n}{k} = 2^n$ ?

Edit

As stated by Hagen Von Elitzen, the result in third question is correct only for square roots of numbers which are pairwise coprime. Eg. ($\sqrt{2}, \sqrt{3}, \sqrt{5}$) and not ($\sqrt{2}, \sqrt{5}, \sqrt{10}$)

Best Answer

Here's a partial answer.

Your idea does indeed work. If you apply it, then you will get the polynomial$$p(x)=x^8-40x^6+353x^4-960x^2+576.$$Asserting that it is a polynomial with the smallest degree within the non-null polynomials of which $\sqrt2+\sqrt3-\sqrt5$ is a root is the same thing as asserting that $p(x)$ is irreducible in $\mathbb{Q}[x]$. Right now, I don't see how to prove it. Now, assume that it is true. Then, if $q(x)\in\mathbb{Q}[x]\setminus\{0\}$ is such that $q\bigl(\sqrt2+\sqrt3-\sqrt5\bigr)=0$, I will prove that the degree of $q(x)$ is greater than or equal to $8$.

Since $p(x)$ and $q(x)$ have a common root, then they are not relatively prime in $\mathbb{Q}[x]$. Therefore, there is a non-constant polynomial $r(x)\in\mathbb{Q}[x]$ that divides both of them. But, since $p(x)$ is irreducible, this means that $r(x)$ can only be of the form $\lambda p(x)$, for some $\lambda\in\mathbb Q$. But then $p(x)$ itself divides $q(x)$ and therefore the degree of $q(x)$ is greater than than or equal to the degree of $p(x)$, which is $8$.

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