Show-that-4-3n-2-5-is-divisible-by-9-then-simplify-4-3n-2-5-9-

Question Number 171572 by Raxreedoroid last updated on 18/Jun/22

Show that

4^{3 n - 2} + 5 is divisible by 9

then simplify \frac{4^{3 n - 2} + 5}{9}

Commented by kaivan.ahmadi last updated on 18/Jun/22

n = 1 \Rightarrow 9 ∣ 4 + 5

n = k \Rightarrow 9 ∣ 4^{3 k - 2} + 5

n = k + 1 \Rightarrow 4^{3 (k + 1) - 2} + 5 = 4^{3 k - 2} \times 4^{3} + 5 =

4^{3} (4^{3 k - 2} + 5) - (4^{3} - 1) \times 5 =

4^{3} (4^{3 k - 2} + 5) - 63 \times 5

w i t h h y p o t h e s i s 9 ∣ 4^{3} (4^{3 k - 2} + 5)

a n d c l e a r l y 9 ∣ 63 \times 5 .

Answered by floor(10²Eta[1]) last updated on 18/Jun/22

for all n \in Z, 3 n - 2 = 3 n + 1 (1 \equiv - 2 (\mod 3))

3 n - 2 : (\dots, - 2, 1, 4, 7, \dots .)

3 n + 1 : (\dots, 1, 4, 7, \dots)

4^{3 n - 2} + 5 = 4^{3 n + 1} + 5 \equiv 64^{n} . 4 + 5 \equiv 4 + 5 \equiv 0 (\mod 9)