Team Selection Test for IMO 2011

The answer is $p^n$. Let $a_i=\frac{(1+p{)}^i-1}{p}$ for $i\ge 0$. By induction we obtain $f(a_i)\equiv if(1)(\mathrm{mod}{p}^n)$ for all $i\ge 0$. If $p>2$ or $p=2$ and $n=1$, then $$a_i\equiv a_j(\mathrm{mod}{p}^n)\Rightarrow (1+p{)}^{j-i}\equiv 1(\mathrm{mod}{p}^{n+1})\Rightarrow i\equiv j(\mathrm{mod}{p}^n).$$ Since $a_i+a_j+p{a}_i{a}_j=a_{i+j}$ for all $i,j\ge 0$, the function defined by $f(a_i)\equiv ic(\mathrm{mod}{p}^n)$, $(0\le i<p^n)$, satisfies the condition of the question for any choice of $c\in {\mathbb{Z}}_{p^n}$.

$$a_i\equiv a_j(\mathrm{mod}{2}^n)\Rightarrow 3^{j-i}\equiv 1(\mathrm{mod}{2}^{n+1})\Rightarrow i\equiv j(\mathrm{mod}{2}^{n-1})$$ and $Z_{p^n}=\{a_i:0\le i<2^{n-1}\}\cup \{-a_i-1:0\le i<2^{n-1}\}$. We also have $2f(-1)\equiv f(-1)+f(-1)\equiv f((-1)+(-1)+2(-1)(-1))\equiv f(0)\equiv 0(\mathrm{mod}{2}^n)$. Since $a_i-a_j+2a_i{a}_j=a_{i+j}$, $a_i+(-a_j-1)+2a_i(-a_j-1)=-a_{i+j}-1$, and $(-a_i-1)+(-a_j-1)+2(-a_i-1)(-a_j-1)=a_{i+j}$ for all $i,j\ge 0$, the function defined by $f(a_i)\equiv ic(\mathrm{mod}{2}^n)$ and $f(-a_i-1)\equiv ic+d(\mathrm{mod}{2}^n)$, ($0\le i<2^{n-1}$), satisfies the condition of the question for any choice of $c\in 2{\mathbb{Z}}_{2^n}$ and $d\in 2^{n-1}{\mathbb{Z}}_{2^n}$ as $f(a)+f(-1)\equiv f(-a-1)(\mathrm{mod}{2}^n)$ for all $a\in Z_{p^n}$.