Problem 1618

Existence: Dividing both sides by $8$, we require
$$251 = \sum_{i=0}^{2005} a_{i+3} 2^i.$$
Now $251 = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7$. Also $-(1+2+\dots+2^{m-1}) + 2^m = 1$, for $m = 1, 2, \dots$ So
$$251 = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7(-1 - 2 - \dots - 2^{1997} + 2^{1998}).$$
So $a_3 = a_5 = a_6 = a_7 = a_8 = a_9 = a_{2008} = +1$ and $a_4 = a_{10} = a_{11} = a_{12} = \dots = a_{2007} = -1$ gives a solution.

Uniqueness: More generally, for $n \ge 1$, each odd integer $m$ with $-2^n < m < 2^n$ has a unique expression as
$$m = \sum_{i=0}^{n-1} a_i 2^i, \quad \text{where } a_i = \pm 1, \text{ for each } i. \qquad (4)$$
We prove this by induction on $n$. The base case $n = 1$ is just the statement that $1 = 2^0$ and $-1 = -2^0$.
Let $n > 1$ and assume the result for $n-1$. Then there is a unique integer $a_0 = \pm 1$ such that $m - a_0 \equiv 2 \pmod 4$. Clearly $-2^n < m - a_0 < 2^n$. So $-2^{n-1} < (m - a_0)/2 < 2^{n-1}$ and $(m - a_0)/2$ is odd. The inductive hypothesis implies that
$$(m - a_0)/2 = \sum_{i=1}^{n-1} a_i 2^{i-1}, \quad \text{for unique } a_i = \pm 1.$$
This gives the expression for $m$. The uniqueness of the expression is apparent from the construction. It is also a consequence of the fact that there are $2^n$ odd integers $m$ with $-2^n < m < 2^n$, but only $2^n$ expressions $\sum_{i=0}^{n-1} a_i 2^i$.