Confusing Myself!

I thought up a problem that I can't solve. Here it is:

Find solutions in integers n and x to the equation n^(x+2) + (n+1)^(x+1) = (n+2)^x

I found one solution, namely (n, x) = (1, 2).

However, for greater n's, the problem becomes nearly impossible to look at with number sense alone. (At any level, it can't be solved algebraically.) I've used Mathematica to determine that there aren't solutions up through n=19, and beyond there x has to be huge. Using the little number theory I know, I determined that all the solutions have to be for odd n. Beyond that, for a specific n, it is possible to find what form x has to take. (For example, when n is 9, x has to be even. This can be determined by looking at the digits value of the powers of 9 and 10, and seeing whether they sum to 1, the digits value of all powers of 11. You end up seeing that the powers of 9 are equal to 1 when x is even, and since the digits value of the powers of 10 is always 0 the sum of the LHS has the same digits value as the RHS, making solutions possible only for those values of x.)

I've also narrowed the solutions down further, and at this point I'm just trying to determine criteria that solutions will have to meet so they can be more easily determined, if they exist at all. (At this point, the outlook on finding more solutions is pretty bleak- I'm feeling pretty confident that there aren't any others- however, I've got to prove it and at the moment that looks to be a pretty difficult task.)

Here's a quick and simple one.

The arithmetic average of 2kn^2 and 2k(n+1)^2 is always k greater than their geometric average for positive integers k and n.

There wasn't a uniform way to find those 2 numbers- I had to do it by finding a bunch that worked and looking for a pattern.

Note: The arithmetic average of 2 numbers a and b is (a+b)/2, and their geometric average is the Sqrt(a*b).

I decided I'd share a cute little power-summation formula I came up with.

Sum[k=0, n-1](x^k) = (x^(n)-1)/(x-1)

I came across it by accident. I noticed that the formula works for the x=2 case, and tried it for x=3. However, I excluded the (x-1) term since I hadn't yet discovered it- for x=2, it disappears. (It's just a 1.) So, I took a shot and added the term- and it worked! Magic!

Nothing special, but it's cool the way the terms cancel. (If you write out the sum and multiply by x-1, you can see that all but the first and last terms cancel out.)