Đặt \(A=p^2+2^p\)
Xét:
+)TH1:p chẵn => p=2
\(\Rightarrow A=2^2+2^2=8\left(ktm\right)\)
+TH2:p lẻ.Nếu p=3k=>p=3
\(\Rightarrow A=3^2+2^3=17\left(tm\right)\)
*Nếu p=3k+1
\(\Rightarrow A=\left(3k+1\right)^2+2^p\)
\(\Rightarrow A=\left(3k+1\right)^2+\left(3-1\right)^p\)
\(\Rightarrow A=9k^2+6k+1+B\left(3\right)-1\)
\(\Rightarrow A=9k^2+6k+B\left(3\right)⋮3\left(ktm\right)\)
*Nếu p=3k+2
(tương tự)
\(\Rightarrow A=9k^2+12k+3+B\left(3\right)⋮3\left(ktm\right)\)
Vậy....