MathDB

B3

Part of 1948 Putnam

Problems(1)

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

Source: Norwegian Mathematical Olympiad 1996 - Abel Competition p2

2/11/2020
Prove that [n+n+1]=[4n+1][\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+1}] for all n∈Nn \in N.
algebrafloor functionfunction