Irish Mathematical Olympiad

a. Let $F_k(n)$ be the collection of sequences/vectors $x$ corresponding to $f_k(n)$. Suppose $x=(x_1,...,x_n)\in F_1(n)$. Then at least one of the entries $x_i\ne 0$. For the first such nonzero value, change its sign and call the resulting sequence $\alpha (x)$. Since $\sum x_i\equiv 1(\mathrm{mod}4)$, we have $\sum \alpha (x)\equiv 3(\mathrm{mod}4)$. The map $\alpha$ is also an involution. Thus $$f_1(n)=f_3(n).(5)$$

b. Applying the same map $\alpha$ to the set $F_0(n)-\bar{0}$ (where $\bar{0}$ is the sequence of all zero's) gives a bijection between $F_0(n)-\bar{0}$ and $F_2(n)$. Thus $$f_0(n)-1=f_2(n).(6)$$ Since these sum residue classes partition all such sequences we have $$f_0(n)+f_1(n)+f_2(n)+f_3(n)=3^n.(7)$$ One more equation is required. Let $A_n=F_0(n)\cup F_2(n)$, the set of all sequences that have an even sum, and let $B_n=F_1(n)\cup F_3(n)$, the set of all sequences that have an odd sum. For a sequence $x=(x_1,\dots ,x_n)$, let $\beta (x)$ be the sequence obtained by finding the first element of the sequence $x$ that is not 1, and changing it with the third entry (i.e. $-1\mapsto 0$ or $0\mapsto -1$). The operation $\beta (x)$ is well defined for $x\ne (1,\dots ,1)$. It also changes the parity of the sum. Thus $\beta :A_n-\bar{1}\mapsto B_n-\bar{1}$ is a bijection. The parity of $n$ tells us which of $A_n$ and $B_n$ the vector $\bar{1}$ is in, hence $$f_0(n)+f_2(n)=f_1(n)+f_3(n)+(-1{)}^n.(8)$$ Combining (1), (2), (3) and (4) give $$f_0(n)=\frac{3^n+2+(-1{)}^n}{4}.(9)$$