I need help on how to prove that 9^(n+3) + 4^n is divisible by 5. Please help, I have no idea of how to solve this
$\displaystyle $9^n 9^3 + 4^n$$
$\displaystyle $(5+4)^n 729 + 4^n$$
Use binomial theorem to expand $\displaystyle $(5+4)^n$$
All terms except the last term $\displaystyle $4^n$$ is divisible by 5.
Take the two left over terms.
$\displaystyle $4^n * 729 + 4^n$$. This is divisible by 5.
even simpler:
9 = 4 (mod 5), whence 9^{3} = 4^{3} = 4 (mod 5) (since 4^{2} = 16 = 1 (mod 5)).
thus 9^{n+3} + 4^{n} = (4^{n})4 + 4^{n} = 5(4^{n}) = 0 (mod 5).
why do this? because why should i have to calculate the cube of 9, when i can calculate the cube of 4 instead (729 is a number i don't use everyday)?