Irish Mathematical Olympiad

a. Because $6^2 \equiv 2 \pmod{17}$, conditions (i) and (ii) imply
$$f(2) \equiv f(6^2) \equiv 2f(6) \pmod{8}$$
If a loggy function $f$ satisfies $f(2) \equiv 1 \pmod{8}$, we get $2f(6) \equiv 1 \pmod{8}$. But the congruence $2x \equiv 1 \pmod{8}$ has no solution $x \in \mathbb{Z}$. Hence, there does not exist a loggy function satisfying $f(2) = 1$.

b. Note that (ii) implies that $f(x^n) \equiv nf(x) \pmod{8}$ for all $x$ that are not divisible by $17$. In particular, $f(3^n) \equiv nf(3) \pmod{8}$. If $f(3) = 1$, this means that for all positive integers $n$ we need to have $f(3^n) \equiv n \pmod{8}$.
We claim that each integer $x$ that is not divisible by $17$ is congruent to $3^n \pmod{17}$ with a unique $1 \le n \le 16$. The reason is that the multiplicative order of $3$ (mod $17$) is equal to $16$. In other words, no two of the numbers $3^n, n = 1, 2, \dots, 16$ are congruent (mod $17$). To see this, suppose $3^a \equiv 3^n \pmod{17}$ for some $1 \le a \le 16$. If $d = \gcd(a, 16)$, there exist positive integers $r, s$ such that either $ar = d + 16s$, or $16s = d + ar$. Because $3^a \equiv 3^{16} \equiv 1 \pmod{17}$ both equations imply $3^d \equiv 1 \pmod{17}$. As $d$ is a factor of $16$, we just check
$$3^2 \equiv 9 \pmod{17}, \quad 3^4 \equiv -4 \pmod{17}, \quad 3^8 \equiv -1 \pmod{17}$$
to see that $d = 16$ and so $a = 16$ as well. Therefore, if $1 \le m \le n \le 16$ and $3^m \equiv 3^n \pmod{17}$, we have $3^{n-m} \equiv 1 \pmod{17}$ and $m \ne n$ would imply $n-m=16$, which is impossible.
Define a function $f: \mathbb{Z} \to \mathbb{Z}$ as follows:
$$f(x) = 0 \quad \text{whenever } x \equiv 0 \pmod{17}$$
$$f(x) = n \quad \text{whenever } x \equiv 3^n \pmod{17} \text{ with } 1 \le n \le 16.$$
Because the value of $f$ depends only on $x$ (mod $17$), condition (i) is satisfied. The function is defined for all integers, because each integer that is not divisible by $17$ is congruent to a unique $3^n \pmod{17}$. Finally, to see that Condition (ii) holds, let $m$ and $n$ both be integers between $1$ and $16$ and note that in case $m+n > 16$, we have $3^{m+n} \equiv 3^{m+n-16} \pmod{17}$ by Fermat's Little Theorem and so
$$f(3^{m+n}) = f(3^{m+n-16}) = m+n-16 \equiv m+n \pmod{8}.$$
Hence, we obtain the required
$$f(3^m \cdot 3^n) = f(3^{m+n}) \equiv m+n \equiv f(3^m) + f(3^n) \pmod{8}.$$
Therefore, this function is a loggy function that satisfies $f(3) = 1$.