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Let n be a positive integer. Prove that: $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$

Martin Sleziak
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Ewin
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1 Answers1

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Case $1$: Let $n=m^2$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m$$ We now have $\sqrt{4n+2} = \sqrt{4m^2+2} \in [2m,2m+1]$. Hence, $\lfloor \sqrt{4n+2} \rfloor = 2m$

Case $2$: Let $n+1=m^2$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m-1$$ We now have $\sqrt{4n+2} = \sqrt{4m^2-2} \in [2m-1,2m]$. Hence, $\lfloor \sqrt{4n+2} \rfloor = 2m-1$

Case $3$: Neither are perfect squares, i.e., $m < \sqrt{n} < \sqrt{n+1} < (m+1)$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m$$ We now have $\sqrt{4n+2} \in \left(\sqrt{4m^2+2},\sqrt{4m^2+8m+2} \right)$. Hence, $\lfloor \sqrt{4n+2} \rfloor = 2m$