# Thread: Induction Proof - Divisibility

1. ## Induction Proof - Divisibility

Could someone check to see if I got this right. It says 'divisible' not 'evenly divisible'. I think it proves divisibility. And is there a difference between 'divisible by ' and 'factor of'? they seem to be the same thing.

$a^{2n}-b^{2n}$ is divisible by $(a+b)$ if $a \ne -b$ . For all positive integers n.

1. first I prove n=1 is true.

$a^{2(1)}-b^{2(1)} = a^{2}-b^{2} = (a+b)(a-b)$

2. then I assume n=k is true and solve for (k+1)th item.

$a^{2(k+1)}-b^{2(k+1)}$

$a^{2k+2}-b^{2k+2}$

$a^{2k+2}-a^{2k}b^{2}+a^{2k}b^{2}-b^{2k+2}$

$a^{2k}(a^{2}-b^{2})+b^{2}(a^{2k}-b^{2k})$

2. Originally Posted by skoker
Could someone check to see if I got this right. It says 'divisible' not 'evenly divisible'. I think it proves divisibility. And is there a difference between 'divisible by ' and 'factor of'? they seem to be the same thing.

$a^{2n}-b^{2n}$ is divisible by $(a+b)$ if $a \ne -b$ . For all positive integers n.

1. first I prove n=1 is true.

$a^{2(1)}-b^{2(1)} = a^{2}-b^{2} = (a+b)(a-b)$

2. then I assume n=k is true and solve for (k+1)th item.

$a^{2(k+1)}-b^{2(k+1)}$

$a^{2k+2}-b^{2k+2}$

$a^{2k+2}-a^{2k}b^{2}+a^{2k}b^{2}-b^{2k+2}$

$a^{2k}(a^{2}-b^{2})+b^{2}(a^{2k}-b^{2k})$

....and?!? You already have it, can you see?

Tonio

3. I am new to doing this kind of proof and have no answer for this problem. I just wanted to make sure this is correct form someone with more skill.

4. Originally Posted by skoker
I am new to doing this kind of proof and have no answer for this problem. I just wanted to make sure this is correct form someone with more skill.
You know that $(a^2-b^2)$ is divisible by $(a+b)$ and by the induction assumption you know that $(a^{2k}-b^{2k})$ is divisible by $(a+b)$

CB

5. ok thanks guys, I just want to make sure I have a clear understanding. and the correct statements etc.