Divisibility by 5

I'm trying to prove that for all integer values of n

9^n - 4^n is a multiple of 5 by induction. I have having trouble that p(n) implies p(n+1). If anyone knows how to do this, any help would be appreciated.

[TomCollins]

9^(n+1) = 9 * 9^n, 4^(n+1) = 4* 4^n,

9^(n+1) - 4^(n+1) = 9* 9^(n) - 9* 4^(n) + 5*4^(n)= 9 * ( some multiple of 5) + 5*(some integer)

[Cooker]

This is simple 9^(n+1)-4^(n+1)=9*9^n-4*4^n=5*9^n+4*(9^n-4^n) which is trivially divisible by 5 if 9^n-4^n is.

[TomCollins]

Let me try to clear this up a bit:
Let 9^n - 4^n = 5*i where i is some integer.

9^(n+1) - 4^(n+1) = 9*(9^n) - 4*(4^n) = 9*(9^n) - 9*(4^n) + 5*(4^n) = 9*(9^n -4^n) + 5* 4^n) = 9*(5*i) + 5*4^n = 5*(9*i + 4^n).

QED.

[Cooker]

Beat me by 3 minutes. I knew I shouldn't have looked over the reasoning that extra time.

thank you

[Cooker]

No problem.

The crux of the proof is that 9=4+5. I believe that it is a trivial extension to further show that for any three natural numbers A,B,C such that A=B+C that C divides A^n - B^n for all n. I think this is totally obvious, but am too lazy to check carefully. This is probably well known, but I am too much of a mathematics novice to be aware of that or not. I did check a few cases and it always worked.

[ QUOTE ]
I believe that it is a trivial extension to further show that for any three natural numbers A,B,C such that A=B+C that C divides A^n - B^n for all n.

[/ QUOTE ]

It's simply because A = B mod C. No induction required.

[TomCollins]

9 = 5 mod 4? Try again.

[AlphaWice]

He means 9 = 4 mod 5.

Look: (5 + 4)^n - 4^n mod 5 is obviously equal to 4^n - 4^n mod 5, which is 0 mod 5.

[BruceZ]

*deleted*

[BruceZ]

[ QUOTE ]
In addition to the other correct solutions, we can use the fact that 9^(n+2) - 4^(n+2) is a difference of squares, and is equal to (9^n - 4^n)*(9^n + 4^n)

[/ QUOTE ]

OOPS! No it's not. I thought of this as I was falling asleep, and was too tired to get up, so I posted it today quickly from memory. Totally bogus, sorry.