đặt \(p^2-q^2-1=a^2\Leftrightarrow p^2-q^2=a^2+1\Leftrightarrow\left(p-q\right)\left(p+q\right)=a^2+1\)
với p,q cùng lẻ \(\Rightarrow\left(p-q\right)\left(p+q\right)\div4\Rightarrow a^2+1⋮4\Rightarrow a^2:4\) dư 3(vô lý)
=> q=2 \(p^2-4-1=a^2\Leftrightarrow\left(p-a\right)\left(p+a\right)=5\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}p-a=1\\p+a=5\end{matrix}\right.\\\left\{{}\begin{matrix}p-a=5\\p+a=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}p=3\\a=2\end{matrix}\right.\\\left\{{}\begin{matrix}p=3\\a=-2\end{matrix}\right.\end{matrix}\right.\)
vậy p=3;q=2
