Where did those operations go?

Trying some values of $n$ , it's obvious that $\left\lfloor \sqrt { 2n+{ n }^{ 2 } } \right\rfloor =n$ , so let's try to prove that. We know that for any real $x$ , $\left\lfloor x \right\rfloor =n \Leftrightarrow n\le x<n+1$ . Let's show that $n\le \sqrt { 2n+{ n }^{ 2 } } <n+1$ . Indeed, since the radical function $\sqrt { . }$ is strictly increasing for positive reals and $0\le { n }^{ 2 }\le { n }^{ 2 }+2n<{ n }^{ 2 }+2n+1={ \left( n+1 \right) }^{ 2 }$ , then $\sqrt { { n }^{ 2 } } \le \sqrt { 2n+{ n }^{ 2 } } <\sqrt { { \left( n+1 \right) }^{ 2 } }$ i.e., $n\le \sqrt { 2n+{ n }^{ 2 } } <n+1$ . Hence $\left\lfloor \sqrt { 2n+{ n }^{ 2 } } \right\rfloor =n$ .