Giả sử tồn tại \(n\in N\) TM.
Ta có: \(n^2+2006=q^2\left(q\in N\right)\)
\(\Leftrightarrow\left(q-n\right)\left(q+n\right)=2006\)\(=2.1003=34.59=17.118=1.2006\)(vì q-n<q+n)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}q-n=2\\q+n=1003\end{matrix}\right.\\\left\{{}\begin{matrix}q-n=34\\q+n=59\end{matrix}\right.\\\left\{{}\begin{matrix}q-n=17\\q+n=118\end{matrix}\right.\\\left\{{}\begin{matrix}q-n=1\\q+n=2006\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}q=\dfrac{1005}{2}\\n=\dfrac{1001}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}q=\dfrac{93}{2}\\n=\dfrac{25}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}q=\dfrac{135}{2}\\n=\dfrac{101}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}q=\dfrac{2007}{2}\\n=\dfrac{2005}{2}\end{matrix}\right.\end{matrix}\right.\)(KTM)
Vậy ko tồn tại số tự nhiên n thỏa mãn.