Question

CMR: n⁶ + n⁴ - 2n² \(⋮\) 72

Answer 1

Ta có : \(n^6+n^4-2n^2=n^2\left(n^2+1\right)\left(n-1\right)\left(n+1\right)\)
\(n=2k\)  hiển nhiên

\(n^2\left(n+2\right)\left(n-1\right)\left(n+1\right)=8k^2\left(n^2+2\right)\left(n-1\right)\left(n+1\right)⋮8\)

\(n=2k+1\) thì \(n^2\left(n+2\right)\left(n-1\right)\left(n+1\right)=n^2\left(n+2\right)\left(2k\right)\left(2k+2\right)\)\(=n^2\left(n^2+2\right)4.k\left(k+1\right)⋮8\)do đó \(k\left(k+1\right)⋮8\) với mọi số nguyên \(m^6+n^{\text{4}}-2n^2⋮8\) (1)

$\star$ $n=3k$ thì ${\mathrm{n2}}^{}({\mathrm{n2}}^{}+2)(n-1)(n+1)=9{\mathrm{k2}}^{}({\mathrm{n2}}^{}+2)(n-1)(n+1)⋮9$

$\star$ $n=3k+1$ thì ${\mathrm{n2}}^{}({\mathrm{n2}}^{}+2)(n-1)(n+1)={\mathrm{n2}}^{}(9k2+6k+3\left)\right(3k\left)\right(n+1)=9{\mathrm{n2}}^{}(3{\mathrm{k2}}^{}+2k+1\left)k\right(n+1)⋮9$

$\star$ $n=3k-1$ thì ${\mathrm{n2}}^{}({\mathrm{n2}}^{}+2)(n-1)(n+1)={\mathrm{n2}}^{}(9{\mathrm{k2}}^{}-6k+3)(n-1)3k=9{\mathrm{n2}}^{}(3{\mathrm{k2}}^{}-2k+1)k(n-1)⋮9$

   $★$ Với mọi $n$ nguyên thì ${\mathrm{n6}}^{}+{\mathrm{n4}}^{}-2{\mathrm{n2}}^{}⋮9$ (2)

từ (1) (2) => đpcm