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: $k+1=2(n+1)$, 
$a^{k+1}-b^{k+1}=a^{2(n+1)} - b^{2(n+1)} = a^{2n}.a^2 - b^{2n}.b^2$
 A: The inductive step is similar to the first problem you did.  Observe
$$(a^{2k} - b^{2k})b^2 = a^{2k} b^2 - b^{2k+2},$$ so
$$(a^{2k} - b^{2k})b^2 + a^{2k+2} - a^{2k} b^2 = a^{2k+2} - b^{2k+2}.$$  Therefore
$$a^{2k+2} - b^{2k+2} = (a^{2k} - b^{2k})b^2 + (a^2 - b^2) a^{2k}.$$
Then by the induction hypothesis, $a+b \mid a^{2k} - b^{2k}$ and we already know $a + b \mid a^2 - b^2$; therefore, $a + b \mid a^{2k+2} - b^{2k+2}$.

If one studies the argument used in the first result, the above computation should be quite natural and does not need to be motivated by repeated induction or more complicated algebra.
In the first result, the inductive step employs the identity $$a^{k+1} - b^{k+1} = a^k (a-b) + (a^k - b^k) b.$$  This takes the $n = k$ case, $a^k - b^k$, multiplies it by $b$ to get $a^k b - b^{k+1}$.  This gives us the $-b^{k+1}$ that we want for the case $n = k+1$, but it is missing $a^{k+1}$ and has an extra $a^k b$; hence the addition of the term $a^k (a-b)$ on the right-hand side.
We do almost exactly the same thing for the second result.  We want to prove the case $n = 2(k+1)$ from the case $n = 2k$, so it is completely natural to take $a^{2k} - b^{2k}$, multiply it by $b^2$ to get $a^{2k}b^2 - b^{2k+2}$, and then note that we need to add $a^{2k+2}$ and subtract $a^{2k}b^2$.  That is all that that is being done in the first part of this answer, and yields a completely analogous identity as in the first result.
For this reason, it is completely perplexing to me how anyone could understand the first result yet fail to comprehend the second.  All we are doing here is establishing a simple algebraic identity.  The identity that was used to show the first result was simply asserted out of nowhere; we could do the same for the analogous identity for the second result.
A: Let $f(x)=x^{2n}-b^{2n}.$
Since $f(\pm b)=0, \ x\mp b \text{ divides } f(x) \text{ in }\mathbb{Z}[x].$
Therefore, $a\pm b \text{ divides } f(a).$
A: Write $a^{2n} - b^{2n}$ as
$$
a^{2n} - b^{2n} = (a^n - b^n)(a^n+b^n).
$$
Using the result you proved (#$1$), $(a+b)$ divides  $a^{2n} - b^{2n}$.
A: 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 $a + b$.
