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[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+1}] , floor function identity

Source: Norwegian Mathematical Olympiad 1996 - Abel Competition p2

February 11, 2020
algebrafloor functionfunction

Problem Statement

Prove that [n+n+1]=[4n+1][\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+1}] for all n∈Nn \in N.
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