<s>[INST] For all $a, b$ in $\mathbb{Z}$, show $a +b$ is a factor of $a^{2n} - b^{2n}$. Given solution approach for first problem below, unable to approach the second one.
#1. For all $a, b$ in $\mathbb{Z}$, show $a - b$ is a factor of $a^n - b^n$.
By weak form of induction.
Base step: $n=1$.
Hypothesis step: Let, $n=k$; then:
$(a-b) \mid (a^k - b^k)$.
Inductive step: $n=k+1$, 
$a^{k+1}-b^{k+1}=a^k(a-b)+(a^k-b^k)b$
Have two terms both divisible by $a-b$.
#2. For all $a, b$ in $\mathbb{Z}$
, show $a + b$ is a factor of $a^{2n} - b^{2n}$.
By weak form of induction.
Base step: $k=1$.
$(a+b) \mid  (a^2 - b^2) = (a+b)\mid (a-b)(a+b)$
Hypothesis step: Let, $k=2n, n \in\mathbb{Z}$; then:
$(a+b) \mid (a^{2n} - b^{2n})$.
Inductive step: $k+1=2(n+1)$, 
$a^{k+1}-b^{k+1}=a^{2(n+1)} - b^{2(n+1)} = a^{2n}.a^2 - b^{2n}.b^2$
 [/INST] Not an answer, but too long for a comment:
You can get Heropup's result by applying the induction step you did in 1) twice. Note that $$\begin{aligned}
a^{2(n + 1)} - b^{2(n + 1)} &= a^{2n + 1}(a - b) + (a^{2n + 1} - b^{2n + 1})b \\
&= a^{2n + 1}(a - b) + (a^{2n}(a - b) + (a^{2n} - b^{2n})b)b
\end{aligned}
$$ Which after simplifying gives
$$
a^{2(n + 1)} - b^{2(n + 1)} = a^{2n + 1}(a - b) + a^{2n}(a - b)b + (a^{2n} - b^{2n})b^2 
$$
The right term $(a^{2n} - b^{2n})b^2$ is divisible by $a + b$ by the induction hypothesis on $a^{2n} - b^{2n}$. For the remaining terms, note that $$
a^{2n + 1}(a - b) + a^{2n}(a-b)b = (a - b)(a^{2n + 1} + a^{2n}b) = (a + b)(a - b)a^{2n},
$$ hence they are also divisible by $a + b$, thus $a^{2(n + 1)} - b^{2(n + 1)}$ is divisible by $a + b$.
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