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

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

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 $n+1$.