Team Selection Test

Let $(b_1,b_2,\dots ,b_k)$ be a permutation of $\{a_1,a_2,\dots ,a_k\}$ which is an arithmetic sequence in (mod $p$). Then, for some integers $c$ and $d$ we have $b_i\equiv c+id(\mathrm{mod}p)$ for every $i=1,2,\dots ,k$.

It is easy to see that $k$ is the order of $a$ in (mod $p$). Then, we have $a^k\equiv 1(\mathrm{mod}p)$ and $$b_1+b_2+\cdots +b_k\equiv a+a^2+\cdots +a^k=\frac{a^k-1}{a-1}\equiv 0(\mathrm{mod}p).$$ On the other hand, we have $$b_1+b_2+\cdots +b_k\equiv \frac{k(2c+d(k+1))}{2}(\mathrm{mod}p)$$ and hence, we get $$2c+d(k+1)\equiv 0(\mathrm{mod}p).$$ Similarly, we have $$b_1^2+b_2^2+\cdots +b_k^2\equiv a^2+a^4+\cdots +a^{2k}=a^2\frac{a^{2k}-1}{a^2-1}\equiv 0(\mathrm{mod}p).$$ since $1<a<p-1$. Moreover, $$\sum _{i=1}^k{b}_i^2=\sum _{i=1}^k(c+id{)}^2=k{c}^2+cdk(k+1)+d^2\frac{k(k+1)(2k+1)}{6}$$ Therefore, we get $$k{c}^2+cdk(k+1)+d^2\frac{k(k+1)(2k+1)}{6}\equiv 0(\mathrm{mod}p)(2)$$ By (1), we can replace $c=-d(k+1)/2$ in (2) and get $$\frac{d^{2}k(k+1{)}^2}{4}-\frac{d^{2}k(k+1{)}^2}{2}+\frac{d^{2}k(k+1)(2k+1)}{6}\equiv 0(\mathrm{mod}p)$$ and hence, $$\frac{d^{2}k(k+1)}{2}\left(-\frac{k+1}{2}+\frac{2k+1}{3}\right)=\frac{d^{2}k(k+1)(k-1)}{12}\equiv 0(\mathrm{mod}p).$$ Clearly, $d\ne 0$ (mod $p$) and $2<k\le p-1$. Therefore, we see that $k=p-1$.