## 17+14sqrt(7) is not the fifth power of an algebraic integer in QQ(sqrt(7))

Prove that $17+14\sqrt{7}$ is not the fifth power of an algebraic integer in $\mathbb{Q}(\sqrt{7})$.

Note that the conjugates of $17+14\sqrt{7}$ are $17 \pm 14\sqrt{7}$, so that the norm of $17+14\sqrt{7}$ in $\mathbb{Q}(\sqrt{7})$ is $17^2 - 7\cdot 14^2 = -1083 = -3 \cdot 19^2$.

Suppose there is an algebraic integer $\zeta \in \mathbb{Q}(\sqrt{7})$ such that $\zeta^5 = 17+14\sqrt{7}$. Then $-3\cdot 19^2 = N(\zeta)^5$; but $N(\zeta)$ is a rational integer, a contradiction. So no such $\zeta$ exists.