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I am working on a homework question that asks to prove that every Integer (including zero) can be calculated in the form $\sum_{k}\pm3^{k}$, where the exponents $k$ are distinct non-negative integers.

I wanted to make sure I am not missing something obvious. What I found was that $\sum_{k}\pm3^{k}\neq0$ when k is distinct. $$3^{0} = 1$$ $$-3^{0} = -1$$ $$-3^{1} + 3^{0} = -2$$ As far as I can tell, there is no way for this expression to equal zero because every other combination will be $\geq 1$ or $\leq -1$. Does this sound correct and if so, what is the correct proof method I can use? I was think proof by counter example.

I found a very similar thread, where sdcvvc pointed out that it was possible if you can represent any number n as $a_{k}3^{k}+a_{k-1}3^{k−1}+⋯+a_{1}3+a_{0}$, such that $a_{i}$∈{−1,0,1}. But without $a_{i}=0$, I don't see how it's possible.

MathProblems...
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In the referenced thread it is demonstrated that every integer $n$ can be written as a sum $$ n = a_{k}3^{k}+a_{k-1}3^{k−1}+ \ldots +a_{1}3+a_{0} $$ where $a_i \in \{ -1, 0, 1 \}$ for $i=0, \ldots, k$. If we denote the indices of non-zero coefficients with $I$ $$ I = \{ i \mid a_i \in \{ -1, 1 \} $$ then $$ n = \sum_{i \in I} a_i 3^i $$ is the representation of $n$ as a sum and difference of distinct powers of $3$.

If $n= 0$ then all $a_i$ are zero and $I$ is the empty set. In this case $\sum_{i \in I} a_i 3^i$ is the empty sum which – by convention – evaluates to zero.

Martin R
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  • I see, thank you. In that case, what would be the best way to prove it? Proof by induction with one case for positive integers and another case for negative integers? – MathProblems... Sep 05 '19 at 15:31
  • @MathProblems...: There is a proof in the Q&A that you referenced, e.g. this one: https://math.stackexchange.com/a/119613/42969. – Martin R Sep 05 '19 at 19:01