Đặt \(M=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(M=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)
\(M=\left(a^2+7a+a+7\right)\left(a^2+5a+3a+15\right)+15\)
\(M=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
Đặt \(p=a^2+8a+11\)
\(\Rightarrow M=\left(p-4\right)\left(p+4\right)+15\)
\(\Rightarrow M=p^2-16+15\)
\(\Rightarrow M=p^2-1\)
\(\Rightarrow M=\left(p-1\right)\left(p+1\right)\)
Thay \(p=a^2+8a+11\)vào M, ta có :
\(M=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)
\(M=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)