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1. \( \left[na\right]=\left[nb\right]\ (\forall)n\in \mathbb{N}\Longleftrightarrow a=b \) (Gh. Andrei)

2. Daca \( a,\ b,\ c,\ d\in\mathbb{N} \) sunt numere consecutive \( \Longrightarrow\left[\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}{2}\right]=\left[\sqrt{a+b+c+d}\right] \) (Dana Piciu)

2) Notam \( a=n,\ b=n+1,\ c=n+2,\ d=n+3 \),

\( x=\frac{\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}}{2},\ y=\sqrt{4n+6} \)

Atunci \( \sqrt{n}+\sqrt{n+3}<\sqrt{4n+6},\ \sqrt{n+1}+\sqrt{n+2}<\sqrt{4n+6} \) deci \( x<y \)

De asemenea se arata usor ca \( \sqrt{4n+5}\le \sqrt{n}+\sqrt{n+3} \) si \( \sqrt{4n+5}\le \sqrt{n+1}+\sqrt{n+2} \), deci \( \sqrt{4n+5}\le x \)

\( \Longrightarrow \sqrt{4n+5}\le x<y \ (1) \) si cum \( 4n+6 \) nu este patrat perfect rezulta ca \( [\sqrt{4n+6}]=[\sqrt{4n+5}] \).

Trecand la partea intreaga in \( (1) \) obtinem ca \( [\sqrt{4n+5}]\le [x]<[\sqrt{4n+6}] \), de unde \( [x]=[y] \).