Can't prove this equation

• December 12th 2012, 04:13 AM
wille13
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
• December 12th 2012, 04:29 AM
MarkFL
Re: Can't prove this equation
I would use induction, observing that:

$(9^{n+4}+4^{n+1})-(9^{n+3}+4^n)=5\cdot9^{n+3}+3(9^{n+3}+4^n)$
• December 12th 2012, 04:29 AM
coolge
Re: Can't prove this equation
$9^n 9^3 + 4^n$
$(5+4)^n 729 + 4^n$
Use binomial theorem to expand $(5+4)^n$
All terms except the last term $4^n$ is divisible by 5.
Take the two left over terms.
$4^n * 729 + 4^n$. This is divisible by 5.
• December 12th 2012, 11:56 AM
richard1234
Re: Can't prove this equation
Or you can observe that

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

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

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

$\equiv 0$ (mod 5)
• December 12th 2012, 04:02 PM
Deveno
Re: Can't prove this equation
Quote:

Originally Posted by richard1234
Or you can observe that

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

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

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

$\equiv 0$ (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)?
• December 12th 2012, 05:18 PM
richard1234
Re: Can't prove this equation
Quote:

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.