We have
$1=3^2-2^3$
$2=3^3-5^2$
$3=2^7-5^3$
$4=5^3-11^2$
$5=3^2-2^2$
and it is unknown if $6$ is representable as a difference of two perfect powers. Next such undecided example is $14$. More: http://oeis.org/A074981
However, I found that
$6=64-49-9=2^6-7^2-3^2$
and
$6=27+4-25=3^3+2^2-5^2$
Similarly
$14=27-9-4=3^3-3^2-2^2$
and
$14=9+9-4=3^2+3^2-2^2$
My question:
Is every positive integer representable in the form: $a_1^{n_1}+a_2^{n_2}-a_3^{n_3}$ or/and in the form $a_1^{n_1}-a_2^{n_2}-a_3^{n_3}$ where $a_1,a_2,a_3,n_1,n_2,n_3$ are natural numbers greater than $1$ with $a_2=0$ also acceptable ?
Are these things known?
The question is based on my own investigation.
There are trivial solutions with squares:
$$n=a^k+b^2-c^2\iff(c+b)(c-b)=a^k-n$$
Choose $a$ to have opposite parity from $n$, so that $a^k-n\pm1$ is even, and then let $c+b=a^k-n$ and $c-b=1$ and solve for $c$ and $b$, i.e.,
$$n=a^k+\left(a^k-n+1\over2\right)^2-\left(a^k-n-1\over2 \right)^2$$
Remark: You could eliminate these trivial solutions by asking that the three powers all be different. Adding such a requirement, though, suggests doing the same for the original problem, and seeing which numbers can be written as a difference of two perfect powers of different degree. So $5=3^2-2^2$ is no longer allowed, but $5=2^5-3^3$ is.