Is $\pi^{2n} + a\pi^n + b$ transcendental, with $a,b$ algebraic?

Question

I got this problem and I've tried something: Suppose $\pi^{2n} + a\pi^n + b = \beta$, $\beta$ algebraic, then

$$\pi^{2n} + a\pi^n + b = \beta \Rightarrow \beta - b = \pi^n(\pi^n + a).$$

I can't proceed, because the product of two transcendental isn't necessarily transcendental. How can I proceed? Thank you!

Answer 1

Complete the square: $\beta - b + a^2/4 = \pi^{2n} + a\pi^n + a^2/4 = (\pi^n + a/2)^2.$ Thus, $\pi^n = \sqrt{\beta - b + a^2/4} - a/2$ and we have our desired contradiction.

Answer 2

Suppose $\beta$ is algebraic. Then $$\pi^{2n} + a \pi^n + b - \beta =0$$ So that $\pi$ is a root of the polynomial $X^{2n} + a X^n + b - \beta$, and is algebraic over the field of algebraic numbers.

But the field of algebraic numbers is algebraically closed. This implies that $\pi$ is algebraic, which is a contradiction.