Question

Chứng minh rằng số A=(n+1)⁴+n⁴+1 chia hết cho 1 số chính phương khác 1 với mọi số n nguyên dương

Answer 1

\(A=\left(n+1\right)^4+n^4+1=\left[\left(n^2+2n+1\right)^2-n^2\right]+\left[\left(n^4+2n^2+1\right)-n^2\right]\)

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

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

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

\(=\left(n^2+n+1\right)\left(2n^2+2n+1\right)=2.\left(n^2+n+1\right)^2⋮\left(n^2+n+1\right)^2\)

\(\Rightarrow A⋮\left(n^2+n+1\right)^2\) => đpcm