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
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.