I'd like to deal with the following problem.
For an integer $n$ greater than $2$, define $u_n = 2^{1/2} +n^{1/n}$ . Prove that there is a polynomial $q_n \in \Bbb Q[x]$ with degrees less than eqaul to $2n$ such that $q_n(u_n) = \sqrt2$
This question asks the Existence of that kind of polynomial.. not construct the polynomial itself specifically.. which means just revealing the existence might be easier than construct one fully. However, can't imagine how to prove the existence of polynomial which satisfies the specific equation.
Any hint/approach to proceed?
HINT:
If $a=\sqrt{2}+\sqrt[n]{n}$ then $$(a-\sqrt{2})^n=n$$ Expand and get $$P(a) -\sqrt{2}Q(a)=0$$ where $P$, $Q$ have integer coefficients, $P$ degree $n$, $Q$ degree $n-1$, $Q$ positive coefficients. From here we get $$(P(a))^2- 2 Q(a)^2=0$$ so an equation of degree $2n$ with integer coefficients for $a$. Back to the previous equality: $$\sqrt{2}=\frac{P(a)}{Q(a)}$$ Now, with some theory of polynomials, one should be able to produce a polynomial expression in $a$ that equals $\sqrt{2}$.