Đặt \(A=7.5^{2n}+12.6^n=7.25^n+12.6^n\)

Do \(25\equiv6\left(mod19\right)\Rightarrow25^n\equiv6^n\left(mod19\right)\)

\(\Rightarrow A\equiv7.6^n+12.6^n\left(mod19\right)\)

\(\Rightarrow A\equiv19.6^n\left(mod19\right)\)

Do \(19.6^n⋮19\Rightarrow A⋮19\)

A = 7.5²ⁿ + 12.6ⁿ

A = 7.(5²)ⁿ + 12.6ⁿ

A = 7.25ⁿ + 12.6ⁿ

25  \(\equiv\) 6 (mod 19)

25ⁿ \(\equiv\) 6ⁿ (mod 19)

7    \(\equiv\) - 12 (mod 19)

⇒ 7.25ⁿ \(\equiv\) -12.6ⁿ (mod 19)

⇒ 7.25ⁿ -( -12.6ⁿ) ⋮ 19

⇒ 7.25ⁿ + 12.6ⁿ   ⋮ 19

Ta có:

\(A=7.5^{2n}+12.6^n=7.25^n+12.6^n\)

Vì \(25\equiv6\left(mod19\right)\Rightarrow25^n\equiv6^n\left(mod19\right)\)

\(\Rightarrow A\equiv7.6^n+12.6^n\left(mod19\right)\)

\(\Rightarrow A\equiv19.6^n\left(mod19\right)\)

\(\Rightarrow A\equiv0\left(mod19\right)\)

Vậy ....

Ta có: \(7.5^{2n}+12.6^n\)

= \(7.5^{2n}+\left(19-7\right).6^n\)

= \(7.5^{2n}+19.6^n-7.6^n\)

= \(7\left(5^{2n}-6^n\right)+19.6^n\)

= \(7\left(25^n-6^n\right)+19.6^n\)

Có: \(19+6^n⋮19\)

\(7\left(25^n-6^n\right)⋮19\)

Vậy...................(đpcm)

Ta có: 3⁴ = 1 (mod 5)

=>3⁴ⁿ = 1ⁿ (mod 5)

=>3⁴ⁿ.3 = 1.3 (mod 5)

=>3⁴ⁿ⁺¹ = 3 (mod 5)

=>3⁴ⁿ⁺¹+2 = 3+2 (mod 5)

=>P = 0 (mod 5)

Vậy P chia hết cho 5(đpcm)

 "=" là đồng dư nha

ta có 3⁴ⁿ⁺¹+2=3⁴ⁿ x 3 + 2= ...1 x 3 +2=...3+2=...5 chia hết cho 5

vậy p chia hết cho 5(đpcm)

P=3⁴ⁿ⁺¹+2

=3⁴ⁿ.3+2

=(3⁴)ⁿ.3+2

=81ⁿ.3+2

=......1.3+2

=.......3+2

=........5 chia hết cho 5 (đpcm)