Question

Chứng minh rằng không tồn tại số nguyên a thỏa mãn \(\left(2017^{2017}+1\right)⋮a^3+11a\)

Answer 1

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

Nếu \(a=3k+1\Rightarrow a^2+11=9k^2+6k+12⋮3\)

Nếu \(a=3k+2\Rightarrow a^2+11=9k^2+12k+15⋮3\)

\(\Rightarrow\left(a^3+11a\right)⋮3\) \(\forall a\in Z\) (1)

Mặt khác ta có:

\(2017\equiv1\left(mod3\right)\Rightarrow2017^{2017}\equiv1\left(mod3\right)\)

\(\Rightarrow\left(2017^{2017}+1\right)\equiv2\left(mod3\right)\)

\(\Rightarrow\left(2017^{2017}+1\right)⋮̸3\) (2)

Từ (1), (2) \(\Rightarrow\left(2017^{2017}+1\right)⋮̸\left(a^3+11a\right)\) \(\forall a\in Z\)