Polya Goes to Work

Not quite sure where we were supposed to be pulling the problem from and no mathematical problem immediately jumps to mind, so I looked back at old exams for one.

Question 1 of the December 2006 exam seems intriguing.

Q: Prove that 4^(n+1) + 5^(2n-1) is a multiple of 21 for all natural numbers n >= 1.

Understanding the problem:
An incoherent and unintuitive mixture of 4 and 5 to some weird combination of exponents gives a multiple of 21.

Plan:
Try to see a pattern. Or just mess around with some induction and hope something pops out. I do like it when a seemingly random mess generates order. Chaos into something intelligible. Scrambled... Ah, off topic.

Carrying it out:
Well, we must begin where the darkness meets the light, where total balance exists, because nothing exists at all - zero.
So, popping in n = 0:
4^(0+1) + 5^(2(0)-1) = 4^1 + 5^(-1) = 4/5.
Eh? Oh wait, yes, that is right, the question says n >= 1... Phew.
n = 1:
4^(1+1) + 5^(2(1)-1) = 4^2 + 5^1 = 21.
Yay, we have our foundation. 21 is a multiple of 21. At least in this universe. And I do exist here. I assumed that when I created this blog.

Now, for some induction.
We assume that for n it's, true, and prove it is true for n+1.

4^(n+1+1) + 5^(2(n+1)-1) = 4^(n+2) + 5^(2n+1)

Now how can I make this so that what we assumed appears within...
So I guess the obvious step is to pull apart the exponents.
Hm...
= 4*4^(n+1) + 5^2*5^(2n-1) Doesn't look good.
We can see what we assumed inside this, but I don't see a way to get at it.
= 4^2*4^n + 5*5^2n Neither does that.
We have 21 from 4^2 + 5 but I don't see a way to work with that either.

It's a exam Q, so it shouldn't be that hard.
Speaking of exams, I need to study for them... Hopefully will finish this later.