$#120914$

`a)` Ta có:
`11 \equiv -4 (mod 15)`
`11^2 \equiv (-4)^2 \equiv 16 \equiv 1 (mod 15)`
`11 ^ (2n) = (11 ^ 2) ^ n equiv 1^n \equiv 1 (mod 15)`
`11^(2n)-1 \equiv 0 (mod 15)`
`11^(2n)-1\vdots 15 (đpcm)`

Vậy...

`b)`

Ta có:

`2^2 = 4 `

`2^(2n) = 4^n`

`to` `2^(2n+1) = 2 * 4^n`

Lại có:

`4 \equiv 4 (mod 9)`

`4^2 \equiv 16 \equiv 7 (mod 9)`

`4^3 = 28 \equiv 1 (mod 9)`

`to`Chu kì là `3: 4^n \equiv {4, 7, 1} (mod 9)`

Có `3 TH`:

Nếu `n \equiv 0: 4^n \equiv 1 ⇒ 2^(2n+1) \equiv 2`

`2 + 12n + 7 \equiv 2 + 0 + 7 = 9 \equiv 0 (mod 9)`

Nếu `n \equiv 1: 4^n \equiv 4 ⇒ 2^(2n+1) \equiv 8`

` 8 + 12n + 7 \equiv 8 + 3 + 7 = 18 \equiv 0 (mod 9)`

Nếu `n \equiv 2: 4^n \equiv 7 ⇒ 2^(2n+1) \equiv 14 \equiv 5`

`5 + 12n + 7 \equiv 5 + 6 + 7 = 18 \equiv 0 (mod 9)`

`to` Trong mọi trường hợp đều `\equiv 0 (mod 9)`

`to` `2^(2n+1) + 12n + 7 \vdots 9` `(đpcm)`

Vậy...