CMR `A= 11^(n+2) + 12^(2n+1) \vdots 133`, với mọi `n \in NN`

`A=11^(n+2)+12^(2n+1)=11^n . 121+12^(2n) . 12=11^n . 121+144^n . 12`

Ta có : `144≡11(mod 133)=>144^n ≡11^n ( mod 133)`

`=>A=11^n . 121+144^n . 12 ≡11^n . 121+11^n . 12=133 . 11^n (mod 133)`

Mà `133. 11^n \vdots 133=>11^n .121+144^n .12\vdots 133`

Hay `A=11^(n+2)+12^(2n+1)\vdots 133AAn\in NN`

Vậy `A=11^(n+2)+12^(2n+1)\vdots 133AAn\in nN`