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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}$.

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    $\begingroup$ $ \forall k \in \mathbb{O}$? $\endgroup$
    – Dietrich Burde
    Mar 28 at 18:40
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    $\begingroup$ Is $\mathbb{O}$ the set of odd integers? $\endgroup$
    – Bruno B
    Mar 28 at 18:45
  • $\begingroup$ It is clearly not correct: $P\implies Q$ does not prove $Q\implies P.$ (Here, $P$ is $\exists a,b,c,d \in \mathbb{N} \cup\{0\}: 3^{a} + k3^{b} = 2^{c} +2^{d}$ and $Q$ is $k\in\mathbb O.$ $\endgroup$
    – Anne Bauval
    Mar 28 at 18:58
  • $\begingroup$ Yes, $\mathbb{O}$ is the set of odd integers. $\endgroup$
    – John Eaton
    Mar 28 at 19:40

1 Answer 1

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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 $k$....

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  • $\begingroup$ How so, I’m curious how you prove this, numerically? $\endgroup$
    – John Eaton
    Mar 28 at 19:22
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    $\begingroup$ For a fixed $k$, one can use bounds for linear forms in logarithms (though some elementary local argument might work too). Generally, the sequences on each side of the equation are very sparse, so one expects their intersection to be small. $\endgroup$
    – Mike Bennett
    Mar 28 at 20:37

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