37th Iranian Mathematical Olympiad

Assume that $p=3$. Then obviously $x_1$ can take any value. Now suppose that $p\ge 5$. We want to prove that $(x_1,x_2,\dots ,x_{\frac{p-1}{2}})\in \{0,1{\}}^{\frac{p-1}{2}}$. Note that $$\sum _{i=1}^{\frac{p-1}{2}}(1-a{x}_i{)}^{\frac{p-1}{2}}\equiv \frac{p-1}{2}+M\left((1-a{)}^{\frac{p-1}{2}}-1\right)(\mathrm{mod}p),$$ where $M$ is common value mod $p$ of $$\sum _{i=1}^{\frac{p-1}{2}}{x}_i^j\left(1\le j\le \frac{p-1}{2}\right).$$ If we choose $a$ such that $(1-a{)}^{\frac{p-1}{2}}=1$, then by using $(1-a{x}_i{)}^{\frac{p-1}{2}}=\pm 1$ or $0$ and the above equation we have for each $i$, $$(1-a{x}_i{)}^{\frac{p-1}{2}}=\left(\frac{1-a{x}_k}{p}\right)=1$$ So if $\left(\frac{t}{p}\right)=1$ we get $$\left(\frac{1-(1-t)x_i}{p}\right)=1$$ Now if $x_i\ne 0$ then the map $f(t)=1-(1-t)x_i$ is bijective. Hence if $P$ is the set of square remainder, $f$ define a bijection from $P$ to itself and $$\sum _{t\in P}f(t)=\sum _{t\in P}t=0,$$ which implies that $x_i=1$.