Problem - 3280
Let $S_n$ be the minimal value of $\displaystyle\sum_{k=1}^n\sqrt{a_k^2+b_k^2}$ where $\{a_k\}$ is an arithmetic sequence whose first term is $4$ and common difference is $8$. $b_1, b_2,\cdots, b_n$ are positive real numbers satisfying $\displaystyle\sum_{k=1}^nb_k=17$. If there exist a positive integer $n$ such that $S_n$ is also an integer, find $n$.
By definition, we have $a_k=4+8(k-1)=8k-4$. $$\therefore\quad\sum_{k=1}^n\sqrt{a_k^2 + b_k^2}=\sum_{k=1}^n\sqrt{(8k-4)^2 + b_k^2}$$
Let $z_k = (8k-4)+ib_k$, then
\begin{align}
\sum_{k=1}^n\sqrt{(8k-4)^2 + b_k^2} & = |4+ib_1| + |12+ib_2| + \cdots + |(8n-4) + ib_n|\\
&\ge |(4+8+\cdots+(8n-4))+i(b_1+b_2+\cdots + b_n)|\\
&= |4n^2 + 17i| \\
&= \sqrt{16n^4+17^2}\\
&= K
\end{align}
where $K$ is an integer.
It follows $$16n^4 + 17^2 = K^2 \implies (K-4n^2)(K+4n^2)=17^2$$
This is a standard indeterminate equation. Because $K+4n^2 > K-4n^2$, it must hold that
$$
\left\{
\begin{array}{ccc}
K+4n^2 &=& 289\\
K-4n^2 &=& 1
\end{array}
\right.
\implies n=\boxed{6}
$$