MathDB

6

Part of 1986 ITAMO

Problems(1)

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

Source: ITAMO 1986 p6

2/2/2020
Show that for any positive integer nn there exists an integer m>1m > 1 such that (2−1)n=m−m−1(\sqrt2-1)^n=\sqrt{m}-\sqrt{m-1}.
diophantineDiophantine equationnumber theory