We can see that \begin{align*} a_1=1\implies a_2=1+1=2\implies a_3=1+1=2\\ a_4=\frac{2}{3}+\frac{3}{2}=\frac{13}{6}=2+\frac{1}{6}\\ [a_4^2]=\bigg[4+\frac{1}{36}+\frac{4}{6}\bigg]=\bigg[4+\frac{25}{36}\bigg]=4 \end{align*}This is true for $n=4$. Assume that, this is true for $n=k$, which imply $[a_k^2]=k$. We need to prove that $[a_{k+1}^2]=k+1$. \begin{align*} a_{k+1}^2=\frac{a_k^2}{k^2}+\frac{k^2}{a_k^2}+2 \end{align*}Assume that $\{a_k^2\}=x$, then \begin{align*} a_{k+1}^2=\frac{k+x}{k^2}+\frac{k^2}{k+x}+2\\ =\frac{k^2+x^2+2kx+k^4}{k^2x+k^3}+2\\ =\frac{k^2+x^2+2kx+k^4+2k^2x+2k^3}{k^2(k+x)}\\ =\frac{x^2+2kx+2k^2x}{k^2(k+x)}+\frac{k^2+2k+1}{k+x}\\ =\frac{\frac{x^2}{k^2}+\frac{2x}{k}+2x}{k+x}+\frac{k^2+2k+1}{k+x} \end{align*}Since $0\leq x<1$, we can see that $k\leq k+x <k+1$, hence \begin{align*} \frac{\frac{x^2}{k^2}+\frac{2x}{k}+2x}{k}+\frac{k^2+2k+1}{k}\geq \frac{\frac{x^2}{k^2}+\frac{2x}{k}+2x}{k+x}+\frac{k^2+2k+1}{k+x}>\frac{\frac{x^2}{k^2}+\frac{2x}{k}+2x}{k+1}+k+1>k+1\\ \frac{1+2k+3k^2}{k^3}+k+2>\frac{\frac{x^2}{k^2}+\frac{2x}{k}+2x+1}{k}+k+2\geq \frac{\frac{x^2}{k^2}+\frac{2x}{k}+2x}{k+x}+\frac{k^2+2k+1}{k+x}> k+1\\ k+2>\bigg[\frac{\frac{x^2}{k^2}+\frac{2x}{k}+2x}{k+x}+\frac{k^2+2k+1}{k+x}\bigg]> k+1 \end{align*}Since that part is an integer, we can say that $[a_{k+1}^2]=k+1$. Therefore, this completes our induction step, thus for all $n\in \mathbb{N}$, we have $[a_n^2]=n$.