# Another Proof via Induction

1. Dec 2, 2005

### GeoMike

The problem asks that I compute $$3^n-2^n$$ for positive integer values of n, starting at 1 and working through a handful of consecutive integers.
From there I am to make a general observation about the values and then attempt to prove it with induction.

So I found the values:
n=1, 3^1-2^1 = 1
n=2, 3^2-2^2 = 5
n=3, 3^3-2^3 = 19
n=4, 3^4-2^4 = 65
n=5, 3^5-2^5 = 211
n=6, 3^6-2^6 = 665
n=7, 3^7-2^7 = 2059
n=8, 3^8-2^8 = 6305
n=9, 3^9-2^9 = 19171

From this I made the observation that for even values of n, $$3^n-2^n$$ is divisible by 5.
So I worked with the idea that $$3^{2n}-2^{2n}$$ is divisible by 5 for ALL positive integers.

Using mathematical induction:

For n=1
$$3^{2(1)}-2^{2(1)} = 5$$

For k
$$3^{2k}-2^{2k}$$

For k+1
$$3^{2(k+1)}-2^{2(k+1)}$$
$$= 3^{2(k+1)}-2^{2(k+1)} -3^{2k} +3^{2k} -2^{2k} +2^{2k}$$
$$= 3^{2k}(3^2-1) - 2^{2k}(2^2-1) + (3^{2k} -2^{2k})$$
The 3rd term is case k, assumed to be divisible by 5

Working with the remaining terms:
$$3^{2k}(8) - 2^{2k}(3)$$
$$= 3^{2k}(8) - 2^{2k}(3) + 2^{2k}(5) - 2^{2k}(5)$$
$$= 3^{2k}(8) - 2^{2k}(8) + 2^{2k}(5)$$
$$= 8(3^{2k} - 2^{2k}) + 5(2^{2k})$$
The first term has case k as a factor, and the second term has 5 as a factor, making both divisible by 5

Thus $$3^{2n}-2^{2n}$$ is divisible by 5 for all positive integers

Does this look valid?
Any way to clean it up?

GM

Last edited: Dec 2, 2005
2. Dec 2, 2005

### HallsofIvy

Staff Emeritus
What you've done is valid but somewhat more complicated than is necessary. Notice that 32(k+1)= 32k+2= 9(32k) and that 22(k+1)= 22k+2= 4(22k). And, of course, 9- 4= 5.

3. Dec 2, 2005

### GeoMike

So:
9(32k) - 4(22k)
5(32k) + 4(32k) - 4(22k)
5(32k) + 4(32k - 22k), both divisible by 5
Yeah, that is much simpler.
I seem to have a gift for missing the obvious and making more work for myself. :tongue:
Thanks again!
GM