Can $n$ be expressed as $a^b-c^d$, where $a,b,c,d,$ and $n$ are natural numbers, not necessary distinct?

Question

Can $n$ be expressed as $a^b-c^d$, where $a,b,c,d,$ and $n$ are natural numbers, not necessary distinct, and $b$ and $d$ can not be both equal to $1$?

For example, when $n=1319$, then $2^{11}-3^6=1319$.

What if $n=2019$?

What if $n=100!$? (! represents the factorial)

For $n=100!$, I do not need to find $a,b,c,$ and $d$. I just want to know if it is possible to be of the given form or no.

Answer 1

Well, if $b$ and $d$ cannot both be equal to $1$ them $2019$ may be rendered as $2025-6=45^2-6^1$. If it is meant that neither $b$ nor $d$ can equal $1$ then you can use a difference of squares, the choice with the smallest numbers for $2019$ being $338^2-335^2$.

All odd numbers and all multiples of $4$ may be rendered as differences of squares.

Answer 2

You can write any multiple of $4$ by using the identity $$(x+1)^2-(x-1)^2=4x$$Since, $100!$ is a multiple of $4$, you can write $$100!=\left(\frac{100!}{4}+1\right)^2-\left(\frac{100!}{4}-1\right)^2$$