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.

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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 $m=a^{2n+1}-a^{2n}b+a^{2n-1}b^2-\dots-a^2b^{2n-1}-ab^{2n}+b^{2n+1}\in\Bbb{Z}.\qquad\square$

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I have two questions:

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

Answer 1

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 $n+1$.

Answer 2

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.

Answer 3

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 $a+b$.

Answer 4

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

Answer 5

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.

Answer 6

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*}