Ireland_2017

a. Because $6^2\equiv 2(\mathrm{mod}17)$, conditions (i) and (ii) imply $$f(2)\equiv f(6^2)\equiv 2f(6)(\mathrm{mod}8)$$ If a loggy function $f$ satisfies $f(2)\equiv 1(\mathrm{mod}8)$, we get $2f(6)\equiv 1(\mathrm{mod}8)$. But the congruence $2x\equiv 1(\mathrm{mod}8)$ 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 $f(x^n)\equiv nf(x)(\mathrm{mod}8)$ for all $x$ that are not divisible by $17$. In particular, $f(3^n)\equiv nf(3)(\mathrm{mod}8)$. If $f(3)=1$, this means that for all positive integers $n$ we need to have $f(3^n)\equiv n(\mathrm{mod}8)$. We claim that each integer $x$ that is not divisible by $17$ is congruent to $3^n(\mathrm{mod}17)$ 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 $3^a\equiv 3^n(\mathrm{mod}17)$ 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 $3^a\equiv 3^{16}\equiv 1(\mathrm{mod}17)$ both equations imply $3^d\equiv 1(\mathrm{mod}17)$. As $d$ is a factor of $16$, we just check $$3^2\equiv 9(\mathrm{mod}17),3^4\equiv -4(\mathrm{mod}17),3^8\equiv -1(\mathrm{mod}17)$$ to see that $d=16$ and so $a=16$ as well. Therefore, if $1\le m\le n\le 16$ and $3^m\equiv 3^n(\mathrm{mod}17)$, we have $3^{n-m}\equiv 1(\mathrm{mod}17)$ and $m\ne n$ would imply $n-m=16$, which is impossible. Define a function $f:\mathbb{Z}\to \mathbb{Z}$ as follows: $$f(x)=0\text{whenever }x\equiv 0(\mathrm{mod}17)$$ $$f(x)=n\text{whenever }x\equiv 3^n(\mathrm{mod}17)\text{ with }1\le n\le 16.$$ 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 $3^n(\mathrm{mod}17)$. 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 $3^{m+n}\equiv 3^{m+n-16}(\mathrm{mod}17)$ by Fermat's Little Theorem and so $$f(3^{m+n})=f(3^{m+n-16})=m+n-16\equiv m+n(\mathrm{mod}8).$$ Hence, we obtain the required $$f(3^m\cdot 3^n)=f(3^{m+n})\equiv m+n\equiv f(3^m)+f(3^n)(\mathrm{mod}8).$$ Therefore, this function is a loggy function that satisfies $f(3)=1$.