<s>[INST] 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.


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?

 [/INST] 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*}
</s>