$(2k+1)^{4k+1} = (2k+1) \pmod {4k+2} $ . . . why? Well, this must be simple but I seem to be dense at the moment. I checked heuristically for small $k$
$$(2k+1)^{4k+1} = (2k+1) \pmod {4k+2} \tag1 $$ 
but couldn't prove that this is so in general.    
For the other quarter of numbers, 
$$(2k)^{4k-1} = 0 \pmod {4k} \tag2 $$
I found at least the (even more simple) solution.    
So: is (1) true? and if, what is the proof?
 A: (1)$(2k+1)^{4k+1}-(2k+1)=(2k+1)((2k+1)^{4k}-1)$ which is clearly divisible by $2k+1$ 
$(2k+1)^{4k}$ is odd, so, $(2k+1)^{4k}-1$  is even.
$\implies gcd(2(2k+1))\mid  ((2k+1)^{4k+1}-(2k+1))$
(2) Let $2^r|| k,$ the highest power of $2$ in $4k$ is $r+2$ and in $(2k)^{4k-1}$ is $(r+1)(4k-1)$ which will be $\ge r+2$  if $k\ge 1$
For other primes $q>2$ , let $q^a||k$ the highest power of $q$ in $4k$ is $a$ and in   $(2k)^{4k-1}$ is $a(4k-1)$  which will be $\ge a$ if $k\ge 1$  
A: For the first one, $(2k+1)^{4k+1} - (2k+1) = (2k+1) \left((2k+1)^{4k} - 1\right)$.
Note that $2 \vert \left((2k+1)^{4k} - 1 \right)$.
Hence, $4k+2 \vert (2k+1)^{4k+1} - (2k+1)$
For the second one. Note that for $k \geq 1$, $(2k)^{4k-1} = 2^3k^3 (2k)^{4k-4}$. Hence, in fact $$(2k)^{4k-1} \equiv 0\pmod{8k^3}$$
A: The number $(2k+1)^{4k+1}-(2k+1)$ is even so 
$$(2k+1)^{4k+1} \equiv (2k+1) \pmod 2  \ .$$
Since the $$(2k+1)^{4k+1} \equiv (2k+1) \pmod {2k+1}  $$
is obvious and $\gcd{(2,2k+1)}=1$ we conclude that 
$$(2k+1)^{4k+1} \equiv (2k+1) \pmod {4k+2} \ . $$
A: Hint $\rm\,\ a,b\:|\:b^n-b\:\Rightarrow\:lcm(a,b)\:|\:b^n-b.\:$ You have $\rm\:a = 2,\ b\:$ odd, so $\rm\:lcm(a,b) = ab.$
Remark $\ $ More generally if $\rm\:a,b\in\Bbb Z,\:$ and $\rm\:f(x)\in \Bbb Z[x]\:$ then
$$\rm a\:|\:f(b),\, b\:|\:f(0)\iff a,b\:|\:f(b)\iff lcm(a,b)\:|\:f(b)$$
Above is the special case $\rm\ f(x) =\, x^n - x.$
