# Thread: Can't prove this equation

1. ## Can't prove this equation

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

2. ## Re: Can't prove this equation

I would use induction, observing that:

$\displaystyle (9^{n+4}+4^{n+1})-(9^{n+3}+4^n)=5\cdot9^{n+3}+3(9^{n+3}+4^n)$

3. ## Re: Can't prove this equation

$\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.

4. ## Re: Can't prove this equation

Or you can observe that

$\displaystyle 9^{n+3} + 4^n = 729(9^n) + 4^n$

$\displaystyle \equiv 729(4^n) + 4^n$ (mod 5)

$\displaystyle \equiv 730(4^n)$ (mod 5)

$\displaystyle \equiv 0$ (mod 5)

5. ## Re: Can't prove this equation

Originally Posted by richard1234
Or you can observe that

$\displaystyle 9^{n+3} + 4^n = 729(9^n) + 4^n$

$\displaystyle \equiv 729(4^n) + 4^n$ (mod 5)

$\displaystyle \equiv 730(4^n)$ (mod 5)

$\displaystyle \equiv 0$ (mod 5)
even simpler:

9 = 4 (mod 5), whence 93 = 43 = 4 (mod 5) (since 42 = 16 = 1 (mod 5)).

thus 9n+3 + 4n = (4n)4 + 4n = 5(4n) = 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)?

6. ## Re: Can't prove this equation

Originally Posted by Deveno
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)?
Yeah that solution's slightly simpler than mine. I just happen to have 9^3 memorized.