75th Romanian Mathematical Olympiad

As $k$ and $p$ are odd, $|K|$ is even, in fact a power of $2$, the characteristic of $K$ being $2$. It follows that $x_1,x_2,\dots ,x_t$ are the roots of $P$. Consider $P={\sum }_{i=0}^t{a}_j{X}^j$.

The multiplicative group $(K^{\ast },\cdot )$ is cyclic of order $kp$. Consider $a\in K^{\ast }$ one of its generators. For any $s\in \{1,2,\dots ,kp\}$, with $(s,kp)=1$ we get $$\mathrm{ord}(a^s)=\mathrm{ord}(a)=kp.(1)$$ In particular, for any $i\in \{1,2,\dots ,p-1\}$ we have $(ik+p,k)=(p,k)=1$ and $(ik+p,p)=(ik,p)=1$, so that $(ik+p,kp)=1$ and $\mathrm{ord}(a^{ik+p})=kp$. These give

$$A_r=\sum _{i=0}^{p-1}{a}^{-irk}P(a^{ik+p})=\sum _{i=0}^{p-1}{a}^{-irk}\sum _{j=0}^t{a}_j{a}^{ikj+pj}=\sum _{j=0}^t{a}_j{a}^{pj}\sum _{i=0}^{p-1}{a}^{k(j-r)i}.$$ Because $$\sum _{i=0}^{p-1}{a}^{kmi}=\{\begin{array}{ll}p,& \text{if }p\mid m,\\ 0,& \text{if }p\nmid m,\end{array}$$ we obtain $$A_r=\sum _{j=0}^t{a}_j{a}^{pj}\sum _{i=0}^{p-1}{a}^{k(j-r)i}=p\cdot \sum _{j=0}^{\lfloor \frac{t-r}{p}\rfloor }{a}_{pj+r}{a}^{p(pj+r)}.$$

For any divisor $d\nmid kp$ consider $U_d=\{x\in K^{\ast }\mid \text{ord}(x)=d\}$ and the polynomial $$Q_d=\prod _{x\in U_d}(X+x).$$ Let $K_2=\{0,1\}$ the prime subfield of $K$ and $1=d_1<d_2<\cdots <d_n=kp$ the divisors of $kp$. As $Q_1=X+1\in K_2[X]$ and $$Q_{d_i}=(X^{d_i}+1)\cdot {\left(\prod _{d_j\mid d_i,d_j\ne d_i}{Q}_{d_j}\right)}^{-1}\in K_2[X],$$ we have that all polynomials $Q_{d_i}\in K_2[X]$ have all coefficients $0$ or $1$. But then $P=(X^{kp}+1)\cdot Q_k^{-1}\in K_2[X]$ has also this property, moreover at least $p$ being non-zero, concluding the proof.