Is 3^n - 2^n Divisible by 5 for Even n?

[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?

Thanks in advance for your time!
GM

 

[HallsofIvy]

What you've done is valid but somewhat more complicated than is necessary. Notice that 3^2(k+1)= 3^2k+2= 9(3^2k) and that 2^2(k+1)= 2^2k+2= 4(2^2k). And, of course, 9- 4= 5.

 

[GeoMike]

What you've done is valid but somewhat more complicated than is necessary. Notice that 3^2(k+1)= 3^2k+2= 9(3^2k) and that 2^2(k+1)= 2^2k+2= 4(2^2k). And, of course, 9- 4= 5.

So:
9(3^2k) - 4(2^2k)
5(3^2k) + 4(3^2k) - 4(2^2k)
5(3^2k) + 4(3^2k - 2^2k), 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.
Thanks again!
GM