67th Czech and Slovak Mathematical Olympiad

Consider the infinite sequence $\{a_n{\}}_{n=1}^{\infty }$ defined by $a_n=\lfloor \sqrt{n}\rfloor$. This sequence is clearly non-decreasing and since $$k=\sqrt{k^2}<\sqrt{k^2+1}<\cdots <\sqrt{k^2+2k}<\sqrt{k^2+2k+1}=k+1,$$ it contains every integer $k$ precisely $(2k+1)$-times. This allows us to express the value of the sum $s_n={\sum }_{i=1}^n{a}_i$ as follows: Let $k=\lfloor \sqrt{n}\rfloor$, that is $n=k^2+l$ for some $l\in \{0,1,\dots ,2k\}$. Then $$\begin{array}{rl}{s}_n& =\sum _{i=0}^{k-1}i(2i+1)+k\cdot (l+1)=2\cdot \sum _{i=1}^{k-1}{i}^2+\sum _{i=1}^{k-1}i+k(l+1)\\ & =2\cdot \frac{(k-1)k(2k-1)}{6}+\frac{(k-1)k}{2}+k(l+1)\\ & =\frac{(k-1)k(4k+1)}{6}+k(l+1),\end{array}$$ where we used $1+2+\cdots +n=\frac{1}{2}n(n+1)$ and $1^2+2^2+\cdots +n^2=\frac{1}{6}n(n+1)(2n+1)$. If $k>6$ then the fraction $\frac{1}{6}(k-1)k(4k+1)$ is an integer sharing a prime factor with $k$, hence the whole right-hand side is sharing a prime factor with $k<s_n$ and $s_n$ is not a prime. If $k\le 6$ then $n<(6+1{)}^2=49$. Plugging $n=48$ into the right-hand side we get $s_{48}=203=7\cdot 29$. For $n=47$ we get $s_{47}=197$, which is a prime. The answer is $n=47$.