prove that $[\sqrt{x}+\sqrt{x+1}]=[\sqrt{4x+2}], x \in \Bbb N$. Could someone help me to solve. I try many ways but I can't solve it ( I tried : Let $x=n^2 +k$ and don't know what to do afterwards).
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1See also: For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$. – Martin Sleziak Feb 04 '19 at 14:11