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(\sqrt2-1)^n=\sqrt{m}-\sqrt{m-1} , for any n exists m>1

Source: ITAMO 1986 p6

February 2, 2020

diophantineDiophantine equationnumber theory

Problem Statement

Show that for any positive integer

n

there exists an integer

m > 1

such that

(\sqrt2-1)^n=\sqrt{m}-\sqrt{m-1}