Ta có:
M = (a + 1)(a + 2)(a + 3)(a + 4) + 1
M = [(a + 1)(a + 4)][(a + 2)(a + 3)] + 1
M = (a2 + 4a + a + 4)(a2 + 3a + 2a + 6) + 1
M = (a2 + 5a + 4)(a2 + 5a + 6) + 1
M = (a2 + 5a + 4)2 + 2(a2 + 5a + 4) + 1
M = (a2 + 5a + 4 + 1)2
M = (a2 + 5a + 5)2
=> M là bình phương của 1 số nguyên
\(M=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)
\(M=\left[\left(a+1\right)\left(a+4\right)\right]\left[\left(a+2\right)\left(a+3\right)\right]+1\)
\(M=\left[a\left(a+4\right)+\left(a+4\right)\right]\left[a\left(a+3\right)+2\left(a+3\right)\right]+1\)
\(M=\left(a^2+4a+a+4\right)\left(a^2+3a+2a+6\right)+1\)
\(M=\left(a^2+5a+5\right)^2-1+1\)
\(M=a^2+5a+5\)
Mà \(a\inℤ\Rightarrow\left(a^2+5a+5\right)\inℤ\)
Vậy,..................
