Question

+Chứng minh:
\(a^7-a\text{ }⋮\text{ }7\text{ }\left(a\in Z\right)\)

Answer 1

A=a^7 -a =a(a^6 -1) =a(a^3 -1)(a^3+1) =(a-1).a.(a+1)[a^2+a+1)(a^2-a+1) ]

\(A=A_0.A_1\)

\(A_1=\left(a^2+a+1\right)\left(a^2-a+1\right)=\left[\left(a^2-4\right)+\left(a+5\right)\right]\left[\left(a^2-9\right)+\left(-a+10\right)\right]\)

\(A_1=\left[\left(a^2-4\right)\left(a^2-9\right)\right]+\left[\left(a^2-4\right)\left(-a+10\right)+\left(a+5\right)\left(a^2-a+1\right)\right]=A_2+A_3\)

\(A_3=\left(a^2-4\right)\left(-a+10\right)+\left(a+5\right)\left(a^2-a+1\right)=-a^3+10a^2+4a-40+a^3-a^2+a+5a^2-5a+5=14a^2-35\)\(A_3=7\left(2a^2-5\right)\)

\(A=A_0.A_1=A_0\left(A_2+A_3\right)=A_0.A_2+A_0.A_3\)

A3 : chia hết cho 7 hiển nhiên => \(A_0.A_3⋮7\)

\(A_0.A_2=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)\left(a^2-9\right)\)

\(A_0A_2=\left(a-3\right)\left(a-2\right)\left(a-1\right)\left(a\right)\left(a+1\right)\left(a+2\right)\left(a+3\right)\)

A0.A2 là tích 7 số nguyên liên tiếp => A0.A2 chia hết cho 7

=>\(A⋮7\) =>dpcm

Answer 2

Ủa cái này là Fermat nhỏ mà.

Answer 3

Cách khác:
Xét \(a\equiv0\left(mod7\right)\)

\(\Rightarrow a^7-a⋮7\)

Xét \(a\equiv1\left(mod7\right)\)

\(\Rightarrow a^7-a\equiv1-1\equiv0\left(mod7\right)\)

Xét \(a\equiv2\left(mod7\right)\)

\(\Rightarrow a^7-a\equiv2.2^6-2\equiv2-2\equiv0\left(mod7\right)\)

......................................................................

\(\Rightarrow a^7-a⋮7\)