What values can $2^j-3^k$ have?
E.g., $$ 2^2-3^1=1\\ 2^2-3^0=3\\ 2^3-3^1=5\\ 2^4-3^2=7 $$
Can every number not divisible by $2$ or $3$ be written as $2^j-3^k$? If not, why?
What values can $2^j-3^k$ have?
5
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number-theory
diophantine-equations
perfect-powers
1 Answers
4
we try to find an $n$ coprime to $6$ such that $2^j-3^k$ doesn't cover all options $\bmod n$.
Since we want $2^j$ to cover a small number of cases we are going to try with $n=2^m-1$.
We find that the order of $3\bmod 511$ is $12$ and the order of $2\bmod 511$ is clearly $9$.
Thus only a small fraction of residues $\bmod 511$ are covered by $2^j-3^k$.
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1... namely, at most $9\cdot12$. +1 – 2017-01-10
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1Using $m=2^{25}-1$ then $2$ has order $25$ and $3$ has order $450$ and with a small heuristic I found only $4122$ different residues of possible $11250 (=25 \cdot 450)$. Nice idea! (+1) – 2017-05-14
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0Thanks, I'm pretty sure I saw you once in the Kassel science department library in the summer of 2013. I think you were reading a yellow book on the Riemann hypothesis. – 2017-05-15