Question

[A B C]

Tìm mọi số tự nhiên n sao cho : 

\(n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 \)  chia hết cho 5

Answer 1

[Đinh quang hiệp]

\(A=n^2+\left(n+1\right)^2+\left(n+2\right)^2+\left(n+3\right)^2=n^2+n^2+2n+1+n^2+4n+4+n^2+6n+9\)

\(=4n^2+12n+14=\left(2n\right)^2+2\cdot2n\cdot3+3^2+5=\left(2n+3\right)^2+5\)

vì \(5⋮5\)để \(A⋮5\Rightarrow\left(2n+3\right)^2⋮5\Rightarrow2n+3⋮5\Rightarrow2n-2+5⋮5\Rightarrow2n-2⋮5\Rightarrow2\left(n-1\right)⋮5\Rightarrow n-1⋮5\)

vì 1 chia 5 dư 1 để n-1 chia hết cho 1 suy ra n chia cho 5 phải dư 1

\(\Rightarrow n=\left(6;11;16;...;5n+1\right)\)

vậy \(n=\left(6;11;16;...;5n+1\right)\)thì \(A⋮5\)