`4n+3 vdots 2n+1`
`=>4n+2+1 vdots 2n+1`
`=>2(2n+1)+1 vdots 2n+1`
`=>1 vdots 2n+1`
`=>2n+1 in Ư(1)={1,-1}`
`*2n+1=1=>2n=0=>n=0(tm)`
`*2n+1=-1=>2n=-2=>n=-1(tm)`
Vậy `n in {0;-1}` thì `4n+3 vdots 2n+1`
\(4n+3⋮2n+1\Leftrightarrow2\left(2n+1\right)+1⋮2n+1\Leftrightarrow1⋮2n+1\)
\(\Rightarrow2n+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
| 2n + 1 | 1 | -1 |
| n | 0 | -1 |
4n + 3 chia hết cho 2n + 1 ( 1 )
Mà 2( 2n + 1 ) chia hết cho 2n + 1 ⇒ 4n + 2 \(⋮\) 2n + 1 ( 2 )
Từ (1) và (2) suy ra: ( 4n + 3 ) - ( 4n + 2 ) chia hết cho 2n + 1 \(\Rightarrow\) 1 \(⋮\) 2n + 1
⇒ 2n + 1 ∈ \(Ư_{\left(1\right)}=\left\{1\right\}\)
2n + 1 =1
2n = 0
⇒ n = 0
