Question

Tìm n thuộc N biết: n^2+(n+1)^2+(n+3)^2 chia hết cho 5

Answer 1

Ta có;

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

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

\(=3n^2+8n+10\)

Ta có:

\(\left[n^2+\left(n+1\right)^2+\left(n+3\right)^2\right]⋮5\)

\(\Leftrightarrow n^2+\left(n+1\right)^2+\left(n+3\right)^2\equiv0\left(mod5\right)\)

\(\Leftrightarrow3n^2+8n+10\equiv0\left(mod5\right)\)

\(\Leftrightarrow3n^2+3n\equiv0\left(mod5\right)\)

\(\Leftrightarrow n\left(n+1\right)\equiv0\left(mod5\right)\)

Do đó n phải có dạng \(5k\) hoặc \(5k+4\)(\(k\in N\))