I posted (or meant to post) essentially the same problem a few weeks ago, here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354033. I will try to put up a solution soon.

Let $ f(x) = \sum_{a \in X} x^a$; the given condition is equivalent to $ f(x) f(x^2) = \frac {1}{1 - x}$, which immediately gives it away: recall that unique binary expansion implies the identity $ \frac {1}{1 - x} = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)...$ so take $ f(x) = (1 + x)(1 + x^4)(1 + x^{16}) ...$. In other words, $ X$ consists of the set of positive integers whose binary expansions in base $ 4$ contain only $ 0$s and $ 1$s. (The bijective proof is straightforward: consider the base-$ 4$ expansion of $ n$ and isolate its digits of $ 2$ and $ 3$, etc.) Agrippina's problem is similar: the identity is $ f(x) f(x^2) f(x^4) = \frac {1}{1 - x}$ and we can take $ f(x) = (1 + x)(1 + x^8)(1 + x^{64})...$.

Ingenious! How did you think of that?

Well, first of all this problem's been posted before (although without the source) and this solution given by several MOPpers, so it wasn't hard to remember. It was also given as a very nice list of examples in this thread. Generally, it is very natural to analyze solutions to equations like $ a + b = n, a \in A, b \in B$ by studying the generating functions of $ A$ and $ B$ because $ AB$ gives all the possible sums at once. For example, the following can be solved by similar means. Putnam 2003 A6: For a set $ S$ of non-negative integers let $ r_S(n)$ denote the number of ordered pairs $ (s_1, s_2) \in S^2, s_1 \neq s_2$ such that $ s_1 + s_2 = n$. Is it possible to partition the non-negative integers into disjoint sets $ A$ and $ B$ such that $ r_A(n) = r_B(n)$ for all $ n$? The important step is to realize that if $ S(x) = \sum_{s \in S} x^s$, then the set of all sums of distinct elements of $ S$ is given by $ S(x)^2 - S(x^2)$ (why?). The rest is computation, and once you've figured out what the answer should be it is not hard to give a direct proof. I'll note that even if you didn't think of binary expansion, repeated application of the problem condition allows you to perform the following calculation: $ f(x) = \frac {1}{(1 - x) f(x^2)} = \frac { f(x^4) (1 - x^2)}{1 - x} = \frac {(1 - x^2)}{(1 - x)(1 - x^4) f(x^4)} = \frac {(1 - x^2)(1 - x^{8}) f(x^{16})}{(1 - x)(1 - x^4)} = ...$ Even if this is not rigorous, it tells you what the answer should look like, but even more it strongly suggests that this is the only answer, not just an answer that works.

My apologies; I misread "integer" as "non-negative integer," and the solution I gave doesn't work as is. I'll keep thinking.

Use induction: Suppose we have integers $ x_1...x_n$ such that every $ x_i + 2x_j$ is distinct. Suppose this set misses the value $ n$. Then add to the set $ k, n-2k$, so now we have $ n$ included as a sum. It is clear that if you take $ k$ arbitrarily large it wont overlap with any previous sums. So such a set exists.

Since nobody has come up with a (complete) solution, let me post mine (inspired by Kalva): First, let us verify that a "base -4" can work. That is, every number can be expressed uniquely as $a_1-4a_2+16a_3-64a_4+\ldots$ where $a_i\in\{0, 1, 2, 3\}$. Uniqueness is evident: if we have $a_1-4a_2+\ldots=b_1-4b_2+\ldots$ then suppose $k$ is the first natural number such that $a_k\neq b_k$. Then $(-4)^{k-1}a_k+(-4)^ka=(-4)^{k-1}b_k+(-4)^kb$ for the rest of the integer $a$ and $b$. Then that means that, dividing by $(-4)^{k-1}$, we have $a_k-b_k=4(a-b)$, so $a_k-b_k$ is divisible by 4, which is impossible because $a_k$ and $b_k$ are not equal and between 0 and 3. So the uniqueness is proved. Now take the $4^k$ numbers $a_1-4a_2+16a_3-\ldots+(-4)^{k-1}a_k$ where $a_i\in\{0, 1, 2, 3\}$. The minimal number possible is $3(-4-64-\ldots)$ and the maximal number is $3(1+16+256+\ldots)$, which cover a range of $3(1+16+256+\ldots)-3(-4-64-\ldots)+1=3(1+4+16+\ldots+4^{k-1})+1=4^k$ numbers. Since there are $4^k$ numbers in the possible range and all numbers are expressed at most once, all numbers must be expressed exactly once. Thus base -4 exists and we can work with it like any normal base. So now as before, we place all numbers with only 0's and 1's in their base -4 expansions in the set. Then for any integer $n$, take its base -4 expansion $n_1-4n_2+16n_3-\ldots$. If $n_i=0$, let $a_i=b_i=0$; if $n_i=1$, let $a_i=1$ and $b_i=0$; if $n_i=2$, let $a_i=0$ and $b_i=1$; and if $n_i=3$, let $a_i=b_i=1$, so that in any case, $a_i+2b_i=n_i$. Then let $a=a_1-4a_2+16a_3-\ldots\in X$ and $b=b_1-4b_2+16b_3-\ldots\in X$ also. Then clearly $a+2b=n$ by construction. We must show uniqueness. For any $a, b\in X$, we have $a-b=(a_1-b_1)-4(a_2-b_2)+16(a_3-b_3)-\ldots$. The first nonzero $a_i-b_i$ is either $1$ or $-1$. Thus $a-b=\pm(-4)^{k-1}+(-4)^kx$ where $x$ is an integer. Hence $|a-b|$ is divisible by $4^{k-1}$ and not $2\cdot4^{k-1}$, so the highest power of two that divides $a-b$ is also a power of 4. Now if $a+2b=c+2d$ for $a, b, c, d\in X$, then $a-c=2(d-b)$. If $a\neq c$ then let us look at the highest power of 2 that divides this common difference. It must be a power of 4 that divides $a-c$, but also is a power of 4 that divides $d-b$ and so is twice a power of 4 that divides $2(d-b)$. No number is both a power of 4 and twice a power of 4, so this contradiction shows uniqueness and we're done.

Using generating functions since they generalize to alot of counting problems, (though this uses neg exponents, but it still is legit) Let $f(x)=(1+x^{1})(1+x^{-4})(1+x^{16})(1+x^{-64})*...(1+x^{(-4)^{a}})$ (the same base -4 thing) Then $f(x)f(x^{2})=\frac{(1+x^{1})(1+x^{2})(1+x^{4})(1+x^{8})....(1+x^{2^{n}})}{x^{r}}$, where $r$ is a pretty huge number, which means at the limiting case, as $a$ tends to infinity, $f(x)f(x^{2})=...+1/x^{3}+1/x^{2}+1/x^{1}+1+x^{1}+x^{2}+x^{3}...$ which solves the problem.

No that's not legit. It's only legit if the generating function converges for some $x$, which this one does not. The generating function approach does not prove the answer, it just points you in the direction. You need another approach to show that it actually works. In the limit, what is $r$? How do you know that the generating function doesn't just become $\ldots+\frac{1}{x^4}+\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1$ and stop there?

Sorry, I was in a rush, and put only the general idea down. It doesn't HAVE to converge for some X, because of the fact that all it's doing is manipulating the exponents of the terms, as a representation of a sumset problem. I'm not ever actually going to evaluate the function $f(x)$. I know it makes absolutely no sense if you considered $x$ as an actual number, but this wasn't the point here. And sorry, I can't calculate right now, but essentially it's pretty easy to show through calculation that the most negative exponent and most positive exponents are increasing arbitrarily large as $a$ approaches infinity.

My point is that you cannot just rush into these calculations. You need the theory of formal power series or (in this case) Laurent series. The calculations you do, taking products of sums and expanding them, is only justified when the power (Laurent) series converges. That's why everyone was up in arms when Euler calculated $\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}$ by setting $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\ldots=x(x^2-\pi^2)(x^2-4\pi^2)\ldots$; he had no rigorous justification for saying two power series were equal just because they had exactly the same roots. And this is a bit like what you're doing; you're saying that the power series $\frac{(1+x)(1+x^2)(1+x^4)\ldots}{x^r}=\ldots+\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1+x+x^2+x^3+\ldots$, whatever $r$ means in the limiting case. Now I'm not saying I don't believe you, because I just gave the same solution set as you. I'm just pointing out that your taking the limit of $f(x)f(x^2)$ as $a$ or $n$ goes to infinity and saying that it equals the product of $\lim_{n\rightarrow\infty}f(x)$ and $\lim_{m\rightarrow\infty}f(x^2)$ is unjustified. It would be justified if it converged for some number $x$, but it doesn't.

Sorry if I was unclear with my words, by limit, I didn't actually mean that the numerical value for some particular $x$, of $f(x)*f(x^2)$; by limit, I just meant that the larger integers (in absolute value) could be represented uniquely once we increased the size of our sequence. I'm not ever using $f(x)*f(x^2)$ as any evaluation for a numerical answer, I'm just using $f(x)*f(x^2)$ conveniently because it just simplified the concept of "looking at all sums". Your example indeed uses these generating functions as actual polynomials, which evaluate numerical answers, which isn't the same as what I'm doing. I can easily just as well put this into an argument without generating functions, but still using the same concept of "the terms acting like a binary string". It's just that the algebraic manipulations are more representative/clear of what the concept is.

zero.destroyer's argument is correct. You don't need convergence for formal power series at all, but it is indeed necessary to be careful with Laurent series' that are infinite in both directions; but this is mostly due to the fact that these do not form a ring, not even a module over the power series' (but over polynomials or finite Laurent sums). The argument here never compares terms in a non-formal nature, unlike Euler, who, as said above, plugged in real numbers to to speak of "roots". It is enough to show that additional factors in the product do not contribute to those summands of exponent $s$, where $|s|$ is bounded by some growing bound dependent on the number of factors. This is the case here. If one wants to do it very formally, an algebraic version would be to state that the module of Laurent series is the projective limit over $n$ of Laurents sums whose exponent's modulus is bounded by $n$. This is just a less comprehensible way to state what I said in the previous paragraph, though. Also, note this "fact": $\sum_{n \in \mathbb Z} x^ n = \sum_{n=1}^ \infty x^{-n} + \sum_{n=0}^ \infty x^n = \frac{x^{-1}}{1-x^ {-1}} + \frac{1}{1-x} = \frac{1}{x-1} + \frac{1}{1-x} = 0$. The error here is that to apply geometric series, you would need it to be a module over power series' (i.e. multiplying any Laurent series with a power series would need to make sense; try to multiply the above one by $\sum_{n=0}^ \infty x^n $ to see the problem). But the problem is not that we lack a common are of convergence for those sums. Actually, the fractions are the meromorphic continuations of the sums (which in turn are the Laurent expansions around $\infty$ and $0$) and as an identity of meromorphic functions, this is completely correct!

Not clear or not if $a$ and $b$ are distinct.

Let $f(X) = \sum_{a \in X} X^a$. The necessary and sufficient condition for $X$ to satisfy the condition is for $$f(X)f(X^2) = \frac 1{1-X}$$for all $x \neq 1$. One can notice that we have $$f(X^4) = \frac 1{(1-X^2)f(X^2)} = \frac 1{(1-X^2) \cdot \frac 1{(1-X)f(X)}} = \frac{f(X)(1-X)}{1-X^2} = \frac{f(X)}{1+X},$$so the infinite product $$\frac{f(X)}{(1+X)(1+X^4)(1+X^{16}) \cdots} = f(X^{2^n}) \to f(0) = 1,$$so the function $$f(X) = (1+X)(1+X^4)(1+X^{16}) \cdots$$can be checked to work. In more concrete terms, we may pick $X$ to be the set of numbers that can be represented as the sum of some distinct nonnegative powers of 4.

We want to find $X$ such that $(\sum_{i\in X} x^i)(\sum_{i\in X} x^{2i})=\dots+x^{-2}+x^{-1}+1+x^1+x^2+\dots$. $(1+x)(1+x^2)=1+x+x^2+x^3$. $(1+x+x^2+x^3)(1+x^{-4}+x^{-8}+x^{-12})=x^{-12}+\dots+x^3$. $(x^{-12}+\dots+x^3)(1+x^{16}+x^{32}+x^{48})=x^{-12}+\dots+x^{51}$. We could keep going like this forever, extending in both directions. Therefore, $X$ is $\{1, -4, 16, -64, \dots\}$ works.

We interpret this using generating functions. Defining $A(x) = \sum_{k \in \mathcal{X}} x^k$, our condition requires \[A(x) A(x^2) = \ldots + x^{-2} + x^{-1} + x^0 + x^1 + x^2 + \ldots.\] to cover all integers exactly once. From here, we note that the functions \begin{align*} A(x) &= \prod \left(1+x^{(-4)^i}\right) = (1+x^1)(1+x^{-4})(1+x^{16})(1+x^{-64}) \ldots \\ A(x^2) &= \prod \left(1+x^{2 \cdot (-4)^i}\right) = (1+x^2)(1+x^{-8})(1+x^{32})(1+x^{-128}) \ldots \end{align*} indeed have the desired product, so our construction for $\mathcal{X}$ is simply \[\boxed{\{\mathcal{X}\} = \text{Integers with only 0 and 1 as digits in base -4}}. \quad \blacksquare\]