Pythogoral sum

Let $a_k=(2k-1)\tan α_k$ . Then $\displaystyle \sum_{k=1}^n (2k-1)\tan α_k=17, \displaystyle \sum_{k=1}^n (2k-1)\sec α_k=S_n$ . So , $\displaystyle \sum_{k=1}^n (2k-1)\sec^2 α_kdα_k=0$ . For $S_n$ to be minimum, $\displaystyle \sum_{k=1}^n \sec α_k\tan α_kdα_k=0$ . Applying LaGrange's method of undetermined multipliers, we get $\displaystyle \sum_{k=1}^n \sec α_k(\sec α_k+\lambda\tan α_k) dα_k\equiv 0$ , where $\lambda$ is an unknown constant to be determined. From this we get $\lambda=-\sin α_k$ for all $k$ , that is, $\sin α_k=$ constant $\implies α_k=α=$ constant. Hence $\displaystyle \sum_{k=1}^n (2k-1)\tan α=17\implies \tan α=\dfrac{17}{n^2}\implies S_n=n^2\sec α=\sqrt {n^4+289}$ . Since $S_n$ has to be an integer, $n^4+289$ must be a perfect square. Let $n^4+289=a^2$ . Then $(n^2+a)(n^2-a)=17\times 17$ . Since $17$ is prime , $n^2-a=1, n^2+a=289\implies n^2=144\implies n=12\implies \dfrac{n}{2}=\boxed 6$ .