$a-b$ divides $a^n-b^n$

Question

I wish to prove that for integers $a,b$ it is always true that $a-b$ divides $a^n-b^n$. I want to do this via induction. For our base case $n=1$ it holds that $a-b|a-b$. Our hypthesis would be that for some $k$ we know $a-b|a^k-b^k$ or equivalently for some integer l: $$a^k-b^k =l (a-b) $$ How would I use this to prove that: $$a^{k+1}-b^{k+1} =l' (a-b) $$

I was thinking I could use some clever factoring or something like the binomial theorem, but not quite sure. Could someone provide a tiny hint?

Any particular reason you want to do this by induction rather than just noting that $$ (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1}) = a^n-b^n $$? — hmakholm left over Monica, Sep 27 '18 at 09:43
The chapter of the book was on induction, so that's why I assumed it could be solved by induction. This is a nice factorisation, though. — , Sep 27 '18 at 17:37
Does this answer your question? Why is $a^n - b^n$ divisible by $a-b$? — user1110341, Aug 17 '23 at 00:32

score 17 · Accepted Answer · answered Sep 27 '18 at 09:42

17

You can use $$a^{k+1}-b^{k+1}=a(a^{k}-b^{k})+b^{k}(a-b).$$

answered Sep 27 '18 at 09:42

studiosus

4,147

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Sometimes the tricks are so simple yet so mindblowing, thanks. – Sep 27 '18 at 09:48
1

That's true but, I'd prefer to do your question as Henning commented. – Mikasa Sep 27 '18 at 10:49

score 3 · Answer 2 · answered Sep 27 '18 at 09:45

3

Add and subtract $ab^n$ from $a^{n+1} - b^{n+1}$.

Note that, $aa^{n} - ab^n + ab^n - bb^{n}$ = $a(a^n - b^n) + (a-b)b^n$. Now you can easily see that this term is divisible by $(a-b)$.

answered Sep 27 '18 at 09:45

Delsilon

316

score 2 · Answer 3 · edited Sep 27 '18 at 09:51

2

Hint:

Try to use Euclid's division algorithm to divide the polynomial $$X^k-1$$ by $X-1$. Do you see a pattern for different $k$'s?

If so, use the induction to prove your hypothesis. You get something of the form $$X^k-1 = (X-1)Q(X)$$ for some polynomial $Q(X)$. Let now $X = \frac{a}{b}, b\neq 0$.

edited Sep 27 '18 at 09:51

Mikasa

67,374

answered Sep 27 '18 at 09:44

abcdef

2,260

score 1 · Answer 4 · answered Nov 19 '20 at 01:06

1

Strong induction may be more natural here.

The base case is $n=1,2$. This is true since $a-b|a-b$ and $a-b|a^2-b^2=(a-b)(a+b)$.

Assume that it's true for $n=1,2,..k$. Then, $$a^{k+1}-b^{k+1}=(a+b)(a^k-b^k)-ab(a^{k-1}-b^{k-1})$$

answered Nov 19 '20 at 01:06

Adola

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$a-b$ divides $a^n-b^n$

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