Why does $n^{n+2}+(n+2)^n\bmod2n+2\equiv0$ for all positive odd integers $n$? Another way of asking:
Why is the expression ${a^b+b^a}\over{a+b}$ an integer whenever $a$ and $b$ are positive odd integers with a difference of 2?
I saw this in a plot of all points $(a,b)$ that make that expression an integer, shown below.

As you can see, the part I am currently interested in is the spiky line that goes from the top left to the bottom right. The center of the line isn't all that interesting ($a=b$), but the regular spikes on the line are. This is where $a$ is odd and $b=a+2$ or vice versa.
I'm just wondering why this is the case?
 A: Since $n$ is odd we can write $n=2m+1$ with $m\in\Bbb N$. Then the statement to prove is
$$4m+4\mid(2m+1)^{2m+3}+(2m+3)^{2m+1}$$
Now
$$(2m+1)^2\equiv4m^2+4m+1\equiv(4m+4)m+1\equiv1\bmod4m+4$$
and
$$(2m+3)^2\equiv4m^2+12m+9\equiv(4m+4)(m+2)+1\equiv1\bmod4m+4$$
Thus $(2m+1)^{2k}\equiv(2m+3)^{2k}\equiv1\bmod4m+4$ for any $k\in\Bbb N$ and
$$(2m+1)^{2m+3}+(2m+3)^{2m+1}\equiv(2m+1)^{2(m+1)}(2m+1)+(2m+3)^{2m}(2m+3)\equiv(2m+1)+(2m+3)\equiv0\bmod4m+4$$
which proves the statement.
A: Since $n\;$is odd,$\;a^n + b^n\;$is divisible by $a+b,\;$for all integers $a,b$.

It follows that$\;n^n + (n+2)^n\;$is divible by $2n+2.\;$Then
\begin{align*}
n^{n+2} + (n+2)^n 
&=\,\left(n^{n+2} - n^n\right)+\left(n^n + (n+2)^n\right)\\[4pt]
&\equiv\, n^{n+2} - n^n\;\;(\text{mod}\;2n+2)\\[4pt]
&\equiv\, n^n(n^2-1)\;\;(\text{mod}\;2n+2)\\[4pt]
&\equiv\, n^n(n-1)(n+1)\;\;(\text{mod}\;2n+2)\\[4pt]
&\equiv\, n^n\left({\small{\frac{n-1}{2}}}\right)(2n+2)\;\;(\text{mod}\;2n+2)\\[4pt]
&\equiv\, 0\;\;(\text{mod}\;\;2n+2)\\[4pt]
\end{align*}
