So, I’ve been trying to determine whether or not this line of reasoning is correct. I’d really like some help.
I want to prove $ \forall k \in \mathbb{O} \exists a,b,c,d \in \mathbb{N} \cup\{0\}: 3^{a} + k3^{b} = 2^{c} +2^{d}$.
I deduced the restriction in $k$ by considering both sides mod $2$.
I.e assuming there exists natural numbers $a,b,c,d$, including 0, such that $3^{a} + k3^{b} = 2^{c} +2^{d} $, we arrive at $ 1+k \equiv$ mod $2$. This reduces to $1 \equiv k\bmod2$. Therefore, $ \forall k \in \mathbb{O} \exists a,b,c,d : 3^{a} + k3^{b} = 2^{c} +2^{d}$.
There are no solutions to $$ 3^a+k 3^b=2^c+2^d $$ with, for example, $k=81,89$ and $97$. Probably, there are no solutions to such equations for almost all $k$....