Question

arata ca (2^n+2 +2^n) | (4^n+1 +4^n) pentru orice nr n

Answer 1

2^n+2+2^n=2^2*2^n+2^n=4*2^n+2^n=5*2^n
4^n+1+4^n=4*4^n+4^n=5*4^n=5*2^2n=5*(2^n)²
5*(2^n)divide 5*(2^n)²  evident

Answer 2

(2ⁿ⁺²+2ⁿ) | (4ⁿ⁺¹ +4ⁿ) , ∀ n

(2ⁿ⁺²+2ⁿ) | (2²)ⁿ⁺¹ +(2²)ⁿ

(2ⁿ⁺²+2ⁿ) | (2²)ⁿ⁺¹ +(2²)ⁿ

(2ⁿ·2²+2ⁿ) | (2²ⁿ·2¹ +2²ⁿ)

(2ⁿ·2²+2ⁿ) | (2ⁿ·2ⁿ·2¹ +2ⁿ 2ⁿ)

2ⁿ·(2²+1) | 2ⁿ·(2ⁿ·2¹ +2ⁿ)

2ⁿ·(4+1) | 2ⁿ·(2ⁿ·2 +2ⁿ)

2ⁿ·5 | 2ⁿ·(2ⁿ·2 +2ⁿ)

2ⁿ I 2ⁿ  ⇒       2ⁿ·5 | 2ⁿ·(2ⁿ·2 +2ⁿ)
 
                      (2ⁿ⁺²+2ⁿ) | (4ⁿ⁺¹ +4ⁿ) , ∀ n