Question

chứng minh 19.8ⁿ + 17 là hợp số ( n thuộc N* , n > 1 )

Answer 1

Giải:

Nếu \(n=2k\)\((k\) \(\in N\)*\()\) thì:

\(19.8^{2k}+17=18.8^{2k}+\left(1+63\right)^k+\left(18-1\right)\)\(\equiv0\) (\(mod\) \(3\))

Nếu \(n=4k+1\) thì:

\(19.8^{4k+1}+17=13.8^{4k+1}+6.8.64^{2k}+17\)

\(=13.8^{4k+1}+39.64^{2k}+9\left(1-65\right)^{2k}+\left(13+4\right)\equiv0\) (\(mod\) \(13\) )

Nếu \(n=4k+3\) thì:

\(19.8^{4k+3}+17=15.8^{4k+3}+4.8^3.64^{2k}+17\)

\(=15.8^{4k+3}+4.510.64^{2k}+4.2\left(1-65\right)^{2k}+\left(25-8\right)\equiv0\) (\(mod\) \(5\))

Vậy \(\forall n\in N\)* \(,n>1\) thì \(19.8^n+17\) là hợp số (Đpcm)