Xét \(a\left(7^a-a^2\right):\)
*C/m: \(a\left(7^a-a^2\right)⋮6\forall a\in N\) (1)
+, \(a\left(7^a-a^2\right)⋮2\)
\(a=2k\left(k\in N\right)\Rightarrow a\left(7^a-a^2\right)⋮2\)
\(a=2k+1\left(k\in N\right)\) .
Ta có: \(7^a\) lẻ mà \(a^2\) cũng lẻ \(\Rightarrow7^a-a^2⋮2\Rightarrow a\left(7^a-a^2\right)⋮2\)
Từ 2 điều trên suy ra (1).
+, \(a\left(7^a-a^2\right)⋮3\) (2)
\(a=3k\left(k\in N\right)\Rightarrow a\left(7^a-a^2\right)⋮3\)
\(a=3k+1\left(k\in N\right)\)
Ta có: \(7^a-a^2=7^a-\left(3k+1\right)^2=7^a-3k\left(3k+2\right)-1=\left(7^a-1\right)-3k\left(3k+2\right)\)
Do \(\left\{{}\begin{matrix}7^a-1⋮\left(7-1\right)=6⋮3\\3k\left(3k+2\right)⋮3\end{matrix}\right.\)\(\Rightarrow\left(7^a-1\right)-3k\left(3k+2\right)⋮3\)
\(a=3k+2\left(k\in N\right)\)
\(7^a-a^2=7^a-\left(3k+2\right)^2=7^a-3k\left(3k+4\right)-4=\left(7^a-1\right)-3k\left(3k+4\right)-3\)
Do \(\left\{{}\begin{matrix}7^a-1⋮\left(7-1\right)=6⋮3\\3k\left(3k+4\right)-3⋮3\end{matrix}\right.\)\(\Rightarrow\left(7^a-1\right)-3k\left(3k+4\right)-3⋮3\)
Từ 3 điều trên suy ra (2)
Từ (1), (2) suy ra \(a\left(7^a+a^2\right)⋮6\)
Tương tự: \(b\left(7^b+b^2\right)⋮6\)
Vậy \(A=a\left(7^a-a^2\right)+b\left(7^b-b^2\right)⋮6\forall a,b\in N\)
