1. Đặt \(16n+1=a^2\) ( \(a\in N\)*, \(a\ge6\) )
\(\Rightarrow16n=a^2-1=\left(a-1\right)\left(a+1\right)\)
Vì \(a\in N\)*, \(a\ge6\) , n nguyên tố nên ta có các TH :
+ TH1: \(\left\{{}\begin{matrix}a-1=8\\a+1=2n\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=9\\2n=10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=5\\a=9\end{matrix}\right.\) (TM)
+ TH2: \(\left\{{}\begin{matrix}a-1=16\\a+1=n\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=18\\a=17\end{matrix}\right.\) (KTM)
+ TH3: \(\left\{{}\begin{matrix}a-1=n\\a+1=16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=15\\n=14\end{matrix}\right.\) (KTM )
+ TH4: \(\left\{{}\begin{matrix}a-1=2n\\a+1=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=7\\n=3\end{matrix}\right.\) ( TM )
bài 2 lm tương tự
