Giả sử \(n^2+17=k^2\) (\(k\in N;k>n\))
\(\Rightarrow k^2-n^2=17\)
\(\Rightarrow\left(k-n\right)\left(k+n\right)=17\)
Do \(\left\{{}\begin{matrix}k-n>0\\k+n>k-n\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k-n=1\\k+n=17\end{matrix}\right.\) \(\Rightarrow n=8\)