I wish to prove that the expression $$ \frac{3^{0}2^{q_0}+3^{1}2^{q_1}+...+3^{k-2}2^{q_{k-2}}+3^{k-1}}{2^{q}-3^k}$$ Can be no positive integer other than 1 for any choice of positive integers $k$ and $q> q_0>q_1>...>q_{k-2}>0$ (note this incidentally imposes $q\ge k$).

For the simple cases of $k=1$ and $k=2$, the result is easy to show. At $k=1$, the numerator is simply $1$, so it is trivial that we cannot obtain an integer other than 1. At $k=2$, the numerator is $2^{q_0}+3$, which is prime for $0<q_0<5$, so we need only consider $q \ge 6$; the inequality $2^q-9 > 2^{q-1}+3 \ge 2^{q_0}+3$, valid for $q>4$, then gives the result. I'd like to generalize this to all positive integer $k$, but I don't see a good way to do this inductively.

One can relate a numerator at a given $k$, call it $n_k$, to the numerator at $k-1$ with the choice of $q_0$ through $q_{k-3}$ each less by $q_{k-2}$, call it $n_{k-1}$, as $n_k=2^{q_{k-2}}n_{k-1}+3^{k-1}$, but I don't see a good way to use this for an inductive argument.


Note that, if we assume the expression is an integer, call it p, for some choice of parameters as above, then we also have the result

$$p=\frac{2^q+3^k}{2^q+3^k}p = \frac{3^{0}2^{q_0+q}+...+3^{k-1}2^q+3^k*2^{q_0}+...+3^{2k-2}2^{q_{k-2}}+3^{2k-1}}{2^{2q}-3^{2k}}$$

Which is an instance of the original expression with $k \rightarrow 2k$ and a new set of $q$'s which satisfy the constraints. We've therefore shown that if there exists a choice of parameters such that the expression is a positive integer other than 1 at some value of $k$, then there also exists a choice of parameters yielding the same integer at $2k$ (incidentally, it's also fairly straightforward to see that one can use the factorization of $2^{qn}-3^{kn}$ to get a similar result for $k \rightarrow nk$ for any positive integer $n$). For this reason, the desired proof is equivalent to showing there exists no working choice of $q$'s for $k$ sufficiently large. That is, it is only necessary to show that $\exists K \in \Bbb{N}$ s.t. there is no working choice of $q$'s $\forall k>K$.



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