5
$\begingroup$

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.

Edit:

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 $\forall k>K$.

$\endgroup$
2

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.