If $\pi$ is transcendental, the $\sqrt{n^2+\pi^2}$ are linearly independant over ${\mathbb Q}$: take a linear combination and notice that $\sqrt{n^2+\pi^2}$ is the only member of the family that is not smooth at $\pi=in$ (this proof would also show independance over ${\mathbb Q}(\pi)$).
GuyR
- 81
- 2