Naturals representable as differences of powers With paper-and-pencil method I found only a first $5$ cases:
$$1=3^2-2^3$$
$$2=3^3-5^2$$
$$3=2^7-5^3$$
$$4=5^3-11^2$$
$$5=2^5-3^3$$
This looks interesting and if a natural $n$ can be represented as difference of two powers (we do not take here $a^1$ into consideration but only exponents $\geq 2$ and we do not take into consideration powers $1^m$) we can call $n$ a power-representable natural number.
It is very reasonable to expect that some numbers can be represented in more than one way but I would like to know here is it known to be true and is it true a following statement:

Every natural number is power-representable.

 A: Differences of squares are well understood.
If
$$
n = a^2 - b^2 = (a-b)(a+b)
$$
then $a-b$ and $a+b$ are either both even or both odd, so $n$ is either odd or a multiple of $4$.
Suppose $n$ is odd. Then each way to write $n = rs$ as a product of two (necessarily odd) factors with $r > s$ tells you
$$
n = \left( \frac{r+ s}{2} \right)^2 - \left( \frac{r- s}{2} \right)^2 .
$$
You can always take $r=n= 2k+1$ and $s=1$, to get the well known
$$
2k+1 = (k+1)^2 - k^2 .
$$
If $n$ is prime that's the only way to write it as a difference of squares.
I leave it to you to find all the ways to write $105 = 3 \times 5 \times 7$.
Then you can work out the argument for the multiples of $4$.
A: A partial answer since as @Ethan points out, the problem is well-known and solved in general. 
I'm going to write an integer of the form $4n$ as a difference of squares. A slightly more subtle argument works for $4n + 1$ and $4n + 3$. 
Suppose that you look at an integer $k = 4n$, where $n$ is a positive integer. $k$ is not prime, for $k = 2(2n)$. If we write 
$$
a + b = 2n
a - b = 2
$$
we have two equations in two unknowns; adding, we get
$$
2a = 2n + 2
$$
so 
$$
a = n + 1
$$
Similarly, $b = n-1$. 
That gives us two integers, $a$ and $b$, with the property that $(a+b)(a-b) = 4n$. But $(a+b)(a-b) = a^2 - b^2$, so our number $k = 4n$ is a difference of squares. 
As an example, look at $k = 4\cdot 5 = 20$, so $n = 5$. The formula above says to pick $a = 6$ and $b = 4$. We compute
$$(a+b)(a-b) = 10 \cdot 2 = 20.$$
 But this is the same as
$$
a^2 - b^2 = 36 - 16 = 20
$$
so we've written $20$ as a difference of squares. 
