1. ## Divisibility

I need help...missed class and i am confused...

For n>1, use congruence theory to establish each of the following divisibility statement

13|3^n+2 + 4^2n+1

2. Hello, terencet!

I have a proof ... hope it's acceptable.

For $\displaystyle n>1$, use congruence theory to establish: .$\displaystyle 13 \,|\,3^{n+2} + 4^{2n+1}$

We have: .$\displaystyle 3^{n+2} + 4^{2n+1} \pmod{13}$

. . . . . .$\displaystyle =\;\left(3^3\right)^{\frac{n+2}{3}} + \left(4^3\right)^{\frac{2n+1}{3}} \pmod{13}$

. . . . . .$\displaystyle = \;(27)^{\frac{n+2}{3}} + (64)^{\frac{2n+1}{3}} \pmod{13}$

. . . . . .$\displaystyle = \;(1)^{\frac{n+2}{3}} + (-1)^{\frac{2n+1}{3}} \pmod{13}$

$\displaystyle \text{The first term is: }\:1^{\frac{n+2}{3}}$
We raise 1 to an integral power and take the cube root.
. . Its value is 1.

$\displaystyle \text{The second term is: }\:(-1)^{\frac{2n+1}{3}}$
We raise -1 to an odd power and take the cube root.
. . Its value is -1.

The expression becomes: .$\displaystyle 1 + (-1) \pmod{13} \quad\Rightarrow\quad 0 \pmod{13}$

Therefore: $\displaystyle 3^{n+2} + 4^{2n+1}$ is divisible by 13.