Question

Chứng minh (n+3)^2-(n-1)^2 chia hết cho 8 với mọi số nguyên

Answer 1

\(\left(n+3\right)^2-\left(n-1\right)^2\)

\(=n^2+6n+9-n^2+2n-1\)

\(=8\left(n+1\right)⋮8\)

Answer 2

Answer 3

\(\left(n+3\right)^2-\left(n-1\right)^2=\left(n+3-n+1\right)\left(n+3+n-1\right)=4\left(2n+2\right)=8\left(n+1\right)⋮8\forall n\in Z\)