Problem - 2072
Let $S_n$ be the minimal value of $\displaystyle\sum_{k=1}^n\sqrt{(2k-1)^2+a_k^2}$, where $n\in\mathbb{N}$, $a_1, a_2, \cdots, a_n\in\mathbb{R}^+$, and $a_1+a_2+\cdots a_n = 17$. If there exists a unique $n$ such that $S_n$ is also an integer, find $n$.
$\displaystyle\sum_{k=1}^n\sqrt{(2k-1)^2+a_k^2}=\displaystyle\sum_{k=1}^n|(2k-1)+i\cdot a_k|\ge |(1+3+\cdots+(2n-1)+i\cdot(a_1+a_2+\cdots+a_n)| =|n^2 + 17i| = \sqrt{n^4 + 17^2} $
Hence $S_n=\sqrt{n^4+17^2}$. Let $n^4 + 17^2 = m^2$, we have $(m+n^2)(m-n^2)=289 \implies \boxed{n=12}$