<s>[INST] Is every integer odd $p$ not divisible by $3$ a sum of difference of powers of $2$ and $3$ or vice versa? Does every odd integer $p$ not divisible by $3$ have the form 
$2^a+3^b$, $2^a-3^b$, and or $3^b-2^a$ for integers $a$ and $b$?
Please show proofs if possible. Thanks for help.
 [/INST] No. $53$ is a counterexample.
There are no integers $a,b$ such that $53=2^a+3^b$ since $$2^a\not=53-3^0,53-3^1,53-3^2,53-3^3$$

Suppose that there are integers $a,b$ such that $53=2^a-3^b$.
It is easy to see that $a\ge 2$ and $b\ge 2$.
We have
$$2\equiv (-1)^a-0\pmod 3\implies \text{$a$ is odd}$$
Also, we have
$$1\equiv 0-(-1)^b\pmod 4\implies \text{$b$ is odd}$$
Here, let $a=2a'+1,b=2b'+1$ where $a',b'$ are positive integers.
Now
$$3\equiv 2\cdot 4^{a'}-3\cdot 9^{b'}\equiv 2(-1)^{a'}-3(-1)^{b'}\equiv 0,1,4\pmod 5$$
which is a contradiction.

Suppose that there are integers $a,b$ such that $53=3^b-2^a$.
It is easy to see that $a\ge 1$ and $b\ge 1$.
We have
$$2\equiv 0-(-1)^a\pmod 3\implies \text{$a$ is even}$$
Also, we have
$$1\equiv (-1)^b-0\pmod 4\implies \text{$b$ is even}$$
So, letting $a=2a',b=2b'$ where $a',b'$ are positive integers, we have
$$3\equiv 9^{b'}-4^{a'}\equiv (-1)^{b'}-(-1)^{a'}\pmod 5\implies \text{$b'$ is odd, $a'$ is even}$$
Letting $a=4k,b=4m+2$ where $k$ is a positive integers and $m$ is a non-negative integer, we have
$$5\equiv 9\cdot 81^m-16^k\equiv 9\cdot 1^m-0\equiv 9\pmod{16}$$
which is a contradiction.
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