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.

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