Originally Posted by

**HTale** Hi guys,

I'm having trouble proving that if $\displaystyle \pi$ is transcendent, then $\displaystyle \pi^{2k}$ and $\displaystyle \beta\pi$ (where $\displaystyle \beta$ is some non-zero rational constant) are transcendental.

I know that if $\displaystyle \beta\pi$ is algebraic, then it is the root of some polynomial, $\displaystyle X-\beta\pi \in \mathbb{Q}[X]$. Then, if this is a linear polynomial in $\displaystyle \mathbb{Q}[X]$, is it right to conclude that any roots will be in $\displaystyle \mathbb{Q}$? If that is the case, can we then set some other polynomial, $\displaystyle \beta X - \beta\pi \in \mathbb{Q}[X]$ and conclude a contradiction?

For the other one, if we have $\displaystyle \pi^{2k}$ is algebraic, then it is the root of $\displaystyle X-\pi^{2k} \in \mathbb{Q}[X]$. I don't know where to proceed from here, really.

Thank you in advance,

HTale.