# Binomial theorem-related proof

1. Sep 13, 2014

### Sheepwall

I've been thinking about this for some time. Now I'm coming on here in the hope of getting some help;

Prove that an+3 + (a + 1)2n+3 is divisible by a2 + a + 1.

I can't quite remember the restrictions on n, though I'd imagine it'd be "for all real n ≥ -1" or something similar.

EDIT: The way to do this would probably be through the method of mathematical induction, since that was the subject of class the day this problem was given.

Last edited: Sep 13, 2014
2. Sep 13, 2014

### Char. Limit

Mathematical induction is definitely the way to go, yeah. Proving that it's true for n=-1 is rather simple, as I'm sure you've already seen. Then you just need to take the expression:

$$a^{(n+1)+3} + (a + 1)^{2(n+1)+3}$$

And rewrite it into an expression of the form:

$$f(a) \left(a^{n+3} + (a+1)^{2n+3}\right)$$

Where f(a) can be any function. Since you know the latter is divisible by $a^2+a+1$ by assumption, then the full expression will be divisible by $a^2+a+1$ as well, and that will conclude the proof!

3. Sep 13, 2014

### Sheepwall

Can I really assume that an+3 + (a + 1)2n+3 is divisible by a2 + a + 1, when that is what I am aiming to prove?

Sorry, just trying to get into the whole induction thinking...

Thanks for your time, btw :) really appreciate it!

4. Sep 13, 2014

### Char. Limit

Yes, you can. In fact, that's the whole point of induction! You prove a certain base case (here, n=-1), then you make the assumption that it's true for an unspecified number n, and using that assumption, prove that it's also true for n+1. Both of those facts together show that it's true for all n≥-1, as long as n's an integer!

5. Sep 13, 2014

### HallsofIvy

No, that's NOT what you are trying to prove. What you are trying to prove is that "an+3 + (a + 1)2n+3 is divisible by a2 + a + 1 for all n". What you are assuming with induction is "an+3 + (a + 1)2n+3 is divisible by a2 + a + 1 for a single value of n.
(I prefer to use a different letter, say "k", rather than "n" when stating induction hypothesis to avoid that mistake: "if a statement is true for n= 1 (or 0 or -1) and whenever the statement is true for a number, k, it I also true for k+1 then the statement is true for all n.")

That is, once you have proved it true for -1, you then show it is true for =1+1= 0. Now that you know it is true for 0, you show it is true for 0+1= 1. Now that you know it is true for 1, you show it is true for 1+ 1= 2.

Showing that "if it is true for k then it is true for k+1" collapses all of those into one proof.

6. Sep 13, 2014

### Sheepwall

Ah, yes, I shall think about that in the future.

As for this problem, though. I still can not solve it :(

7. Sep 13, 2014

### Sheepwall

I'm going to need some hint, I've tried some different ways; Writing (a + 1)^(2n+3) as a sum, futile long division.

8. Sep 24, 2014

### bshoor

In = an+3 + (a+1)2n+3 , when n=-1 and n=0 it can be shown that it is divisible by a2+a+1
In+1 = an+4 + (a+1)2n+5
In+1 - In = (an+4 - an+3) + [(a+1)2n+5 - (a+1)2n+3]
which can be ultimately reduce to In+1 = a.In + (a2+a+1).(a+1)2n+3
So, by induction method u can prove. Not true for n less than -1.