Irish Mathematical Olympiad

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 $-(1+2+\cdots +2^{m-1})+2^m=1$, for $m=1,2,\dots$ So $$251=2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7(-1-2-\cdots -2^{1997}+2^{1998}).$$ So $a_3=a_5=a_6=a_7=a_8=a_9=a_{2008}=+1$ and $a_4=a_{10}=a_{11}=a_{12}=\cdots =a_{2007}=-1$ 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 $$m=\sum _{i=0}^{n-1}{a}_i{2}^i,\text{where }{a}_i=\pm 1,\text{ for each }i.(4)$$ 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 $m-a_0\equiv 2(\mathrm{mod}4)$. 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 $$(m-a_0)/2=\sum _{i=1}^{n-1}{a}_i{2}^{i-1},\text{for unique }{a}_i=\pm 1.$$ 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 ${\sum }_{i=0}^{n-1}{a}_i{2}^i$.