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

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- Dec 12th 2012, 03:13 AMwille13Can'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

- Dec 12th 2012, 03:29 AMMarkFLRe: Can't prove this equation
I would use induction, observing that:

$\displaystyle (9^{n+4}+4^{n+1})-(9^{n+3}+4^n)=5\cdot9^{n+3}+3(9^{n+3}+4^n)$ - Dec 12th 2012, 03:29 AMcoolgeRe: 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. - Dec 12th 2012, 10:56 AMrichard1234Re: 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) - Dec 12th 2012, 03:02 PMDevenoRe: Can't prove this equation
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)? - Dec 12th 2012, 04:18 PMrichard1234Re: Can't prove this equation