a) \(n^2+8n+29=n^2+4n+4n+16+15=\left(n+4\right)^2+15=m^2\)
\(\Leftrightarrow m^2-\left(n+4\right)^2=15\Leftrightarrow\left(m-n-4\right)\left(m+n+4\right)=13=1.13\)
Do \(m-n-4< m+n+4\)nên ta có trường hợp:
\(\hept{\begin{cases}m-n-4=1\\m+n+4=13\end{cases}}\Leftrightarrow\hept{\begin{cases}m=7\\n=2\end{cases}}\)(thỏa)
b) \(9n^2+6n+22=3\left(3n^2+n\right)+3n+1+21=\left(3n+1\right)^2+21=m^2\)
\(\Leftrightarrow m^2-\left(3n+1\right)^2=21\Leftrightarrow\left(m-3n-1\right)\left(m+3n+1\right)=21=1.21=3.7\)
Ta có các trường hợp:
- \(\hept{\begin{cases}m-3n-1=1\\m+3n+1=21\end{cases}}\Leftrightarrow\hept{\begin{cases}m=11\\n=3\end{cases}}\)(thỏa)
- \(\hept{\begin{cases}m-3n-1=3\\m+3n+1=7\end{cases}}\Leftrightarrow\hept{\begin{cases}m=5\\n=\frac{1}{3}\end{cases}}\)(loại)
