Question

Chứng minh rằng vs mọi số nguyên a:

a, a⁴ + 6a³ +11a² + 6a \(⋮\) 24

b, a⁵ - 5a³ + 4a \(⋮\) 120

c, 3a⁴ - 14a³ + 21a² - 10a \(⋮\) 24

\(⋮\) \(⋮\)\(⋮\)

Answer 1

a: \(a^4+6a^3+11a^2+6a\)

\(=a\left(a^3+6a^2+11a+6\right)\)

\(=a\left(a^3+a^2+5a^2+5a+6a+6\right)\)

\(=a\left(a+1\right)\left(a^2+5a+6\right)\)

\(=a\left(a+1\right)\left(a+2\right)\left(a+3\right)\)

Vì a;a+1;a+2;a+3 là bốn số liên tiếp

nên \(a\left(a+1\right)\left(a+2\right)\left(a+3\right)⋮4!\)

hay \(a\left(a+1\right)\left(a+2\right)\left(a+3\right)⋮24\)

b: \(a^5-5a^3+4a\)

\(=a\left(a^4-5a^2+4\right)\)

\(=a\left(a^2-4\right)\left(a^2-1\right)\)

\(=a\left(a-2\right)\left(a+2\right)\left(a-1\right)\left(a+1\right)\)

Vì a;a-2;a+2;a-1;a+1 là 5 số liên tiếp

nên \(a\left(a-2\right)\left(a+2\right)\left(a-1\right)\left(a+1\right)⋮5!\)

hay \(a\left(a-2\right)\left(a+2\right)\left(a-1\right)\left(a+1\right)⋮120\)