Preselection tests for the full-time training

We have $a_{n+1}^2-a_n^2=1$ for all natural number $n$. Using a telescopic sum we obtain $$n_0=\sum _{k=n_0}^{2n_0-1}\left(a_{k+1}^2-a_k^2\right)=a_{2n_0}^2-a_{n_0}^2=9a_{n_0}^2-a_{n_0}^2=8a_{n_0}^2.$$ Hence $$a_{n_0}^2=\frac{n_0}{8}$$ On the other hand, $$n_0-1=\sum _{k=1}^{n_0-1}\left(a_{k+1}^2-a_k^2\right)=a_{n_0}^2-a_1^2=\frac{n_0}{8}-a_1^2,$$ or equivalently $$a_1^2=\frac{8-7n_0}{8}$$ Since $a_1^2\ge 0$, we deduce that $n_0=1$ and $a_1^2=\frac{1}{8}$. Therefore $$45=\sum _{k=1}^{45}\left(a_{k+1}^2-a_k^2\right)=a_{46}^2-a_1^2=a_{46}^2-\frac{1}{8},$$ which leads to $$a_{46}=\frac{19\sqrt{2}}{4}.$$