Prove that $a^{2n} -b^{2n}$ is divisible by $a+b$

Question

Prove that for all $n\in\Bbb{Z}$, $a^{2n}-b^{2n}$ is divisible by $a+b$ using induction.

I know that if $a$ is divisible by $b$, then $a=kb$, where $k\in\Bbb{Z}$. Here we have that $a^{2n}-b^{2n}=(a+b)k$, with $k\in\Bbb{Z}$.

For the base case I set $n=1$, so $a^2-b^2=(a+b)(a-b)=(a+b)k$, where $k=a-b\in\Bbb{Z}$.

Now the inductive step (where I have doubts): $$a^{2n}-b^{2n}=(a+b)k\implies a^{2(n+1)}-b^{2(n+1)}=(a+b)m,\;k,m\in\Bbb{Z}.$$ We start from $a^{2(n+1)}-b^{2(n+1)}$. Then $$a^{2n+2}-b^{2n+2}=(a+b)\color{red}{(a^{2n+1}-a^{2n}b+a^{2n-1}b^2-\dots-a^2b^{2n-1}-ab^{2n}+b^{2n+1})},$$ so $a^{2(n+1)}-b^{2(n+1)}=(a+b)m$, where $m=a^{2n+1}-a^{2n}b+a^{2n-1}b^2-\dots-a^2b^{2n-1}-ab^{2n}+b^{2n+1}\in\Bbb{Z}.\qquad\square$

I have two questions:

Is the math in $\color{red}{\text{red}}$ a correct descomposition of $a^{2(n+1)}-b^{2(n+1)}$?
We have not used the inductive hypothesis. Could we use it?

There is a known trick to rewrite differences of the kind $x^n-y^n$:
$$ x^n-y^n=(x-y)\sum_{k=1}^n x^{k-1} y^{n-k}$$ This will simplify things, I guess. This identity is very useful, so try to remember it ;) — Masacroso, Jul 01 '19 at 03:01
To know that the decomposition is red is true you must assume that the pattern repeats through the "....". Such an assumption is in essence induction. — fleablood, Jul 01 '19 at 03:03
@Masacroso omg sorry for not being active in rinconmatematico!! :(. And thanks for the formula, I hate to remember the combinatorial number :P. I really appreciate it. — manooooh, Jul 01 '19 at 03:03
@fleablood thanks for the help, but I tried to make the product and did not came up with a nice expression (the terms that are not $a^{2n+2}$ nor $b^{2n+2}$ are not canceled each other, so I thought my descomposition failed). — manooooh, Jul 01 '19 at 03:06
Possible duplicate of Why is $a^n - b^n$ divisible by $a-b$?. Use that to show $a^2-b^2\mid a^{2n}-b^{2n}$ and combine with $a+b\mid a^2-b^2$. Downvotes to "trusted" users who failed to search. — Jyrki Lahtonen, Jul 01 '19 at 14:45

score 3 · Accepted Answer · answered Jul 01 '19 at 02:58

The descomposition in red is correct. You did not use it because you could try this without induction, just with the factorization you used above. But you could have used it in the following way:

Since $$\begin{align} a^{2n+2}-b^{2n+2}=& a^{2n+2}-a^2b^{2n}+a^2b^{2n}-b^{2n+2} \\ =& a^2 (a^{2n}-b^{2n})+b^{2n}(a^2-b^2) \\ =& a^2 (a+b)k+b^{2n}(a-b)(a+b) \\ =&(a+b)\cdots \end{align}$$ The assert is even true for $n+1$.

score 3 · Answer 2 · answered Jul 01 '19 at 02:59

3

For your induction step use

$$ a^{2n+2} - b^{2n+2} = a^2(a^{2n}-b^{2n})+b^{2n}(a^2-b^2)$$ and use the induction hypotheses.

answered Jul 01 '19 at 02:59

Mohammad Riazi-Kermani

68,728

Do you search for duplicates at all? – Jyrki Lahtonen Jul 01 '19 at 14:49
Why? Probably I should. – Mohammad Riazi-Kermani Jul 01 '19 at 14:58
Questions that have already been asked and answered should not be allowed to reappear. Having duplicates around makes it more difficult to use the site. True, the askers should do their part, but more experienced users should be role models. – Jyrki Lahtonen Jul 01 '19 at 15:03
Thanks, for the informative and helpful comment. – Mohammad Riazi-Kermani Jul 01 '19 at 15:34
There are differences of opinions about how to best use the site. Check out our meta site for the details. Also, it is more efficient to discuss the norms there. – Jyrki Lahtonen Jul 01 '19 at 15:36

Explorer · Answer 3 · 2019-07-01T03:25:53.363

Answer 1

This can be verfied by noting that the red term is a geometric progression.

Answer 2

As you have rightly mentioned, your proof is not inductive because the inductive hypothesis is not used. The following is the proof by induction: Assume that the statement is true for all integers $\leq n$.

\begin{align} a^{2n+2}-b^{2n+2} &=a^2 (a^{2n}-b^{2n})+a^2b^{2n}-a^{2 n}b^2+b^{2}(a^{2n}-b^{2n})\\ &=(a^2+b^2) (a^{2n}-b^{2n})-a^2b^2(a^{2(n-1)}-b^{2 (n-1)}), \end{align} where both terms are divisible by $a+b$.

Z Ahmed · Answer 4 · 2019-07-01T08:50:46.363

2

This is true by Remainder theorem as the remainder of $a^{2n}-b^{2n}$ by putting $a=-b$ is zero.

edited Jul 01 '19 at 08:50

answered Jul 01 '19 at 06:13

Z Ahmed

43,235

Since the question asking to prove $a + b$ is a factor, don't you mean putting $a = -b$ in instead of $a = b$? – John Omielan Jul 01 '19 at 07:02
@John Omielan Oh! Yes, thanks, I have edited it now. – Z Ahmed Jul 01 '19 at 08:52

nonuser · Answer 5 · 2019-07-01T11:17:12.177

1

Base $n=1$ $$a^2-b^2 = (a-b)(a+b) $$

By induction hypothetis we have $$a^{2n}\equiv b^{2n}\pmod {a+b}$$

From the base case we have also $a^2\equiv b^2 \pmod {a+b}$, so: $$a^{2n+2}= a^2\cdot a^{2n}\equiv b^2 \cdot b^{2n} = b^{2n+2}\pmod {a+b}$$

and we are done.

edited Jul 01 '19 at 11:17

answered Jul 01 '19 at 08:59

nonuser

90,026

Do you search for duplicates at all? – Jyrki Lahtonen Jul 01 '19 at 14:49
Why? Am I a moderator? @JyrkiLahtonen – nonuser Jul 01 '19 at 15:09
All the users (askers and answerers alike) are expected to search for duplicates before posting. The site is community moderated. In this sense as well as all the others. Experienced users are supposed to be role models. – Jyrki Lahtonen Jul 01 '19 at 15:34

score 0 · Answer 6 · answered Jul 01 '19 at 14:37

Another way:

We know by inductive hypothesis $a^{2n}=(a+b)k+b^{2n}$, with $k\in\Bbb{Z}$. Then, \begin{align*}a^{2n+2}-b^{2n+2}&=a^{2n}a^2-b^{2n}b^2\\ &=a^{2}((a+b)k+b^{2n})-b^{2n}b^2\\ &=a^{2}(a+b)k+a^{2}b^{2n}-b^{2n}b^2\\ &=a^{2}(a+b)k+b^{2n}(a^2-b^2)\\ &=a^{2}(a+b)k+b^{2n}(a-b)(a+b)\\ &=(a+b)(a^2k+b^{2n}(a-b))\\ &=(a+b)m,\quad m=a^2k+b^{2n}(a-b)\in\Bbb{Z}.\end{align*}

Prove that $a^{2n} -b^{2n}$ is divisible by $a+b$

6 Answers6