Question

CM rằng :

n^2+4n+3 chia hết cho 8 với mọi n lẻ.

Giúp mình với nha.thanks trước ạ😘

Answer 1

\(n^2+4n+3\\ =n^2+n+3n+3\\ =n\left(n+1\right)+3\left(n+1\right)\\ =\left(n+1\right)\left(n+3\right)\\ n=2k+1\left(k\in Z\right)\\ \Rightarrow n^2+4n+3\\ =\left(2k+2\right)\left(2k+4\right)\\ =2\left(k+1\right)\cdot2\left(k+2\right)\\ =4\left(k+1\right)\left(k+2\right)\)

\(\left(k+1\right)\left(k+2\right)\) là tích hai số nguyên liên tiếp \(\Rightarrow\left(k+1\right)\left(k+2\right)⋮2\Rightarrow\left(k+1\right)\left(k+2\right)=2a\)

\(\Rightarrow n^2+4n+3=8a⋮8\)