ta có \(\left\{{}\begin{matrix}n+20=a^2\\n-69=b^2\end{matrix}\right.\left(n>69\right)\)
\(\Leftrightarrow\left(n+20\right)-\left(n-69\right)=a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=89=1.89\)
vì a-b<a+b nên a-b=1và a+b=89
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a+b=89\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=45\\b=44\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n+20=45^2\\n-69=44^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=2005\\n=1\left(KTM\right)\end{matrix}\right.\)
vậy n=2005
