Problems from Ukrainian Authors

Since the polynomials $P_1,\dots ,P_n$ do not have integer roots then $P_i(0)\ne 0$ for any $i=\overline{1,n}$ and $P_i(0)=\pm 1$. Let us consider all pairs of polynomials $P_k$ and $P_j$, such that $k\ne j$ and $P_k(0)=P_j(0)$. For each pair of polynomials we determine $Q_{k,j}=P_k-P_j$. If $a$ and $b$ are the number of polynomials such that $P_i(0)=1$ and $P_i(0)=-1$ respectively, then the number of polynomials $Q_{k,j}$ will be $$2C_a^2+2C_b^2=a(a-1)+b(b-1)=a^2+b^2-n\ge \frac{1}{2}(a+b{)}^2-n=\frac{1}{2}n(n-2).$$ Note that coefficients of polynomials $Q_{k,j}$ may be only equal to $-2,-1,0,1,2$. Hence, the coefficients of $Q_{k,j}-Q_{i,l}$ cannot be bigger than $4$ by absolute value. Therefore $Q_{k,j}(5)=Q_{i,l}(5)$ if and only if $Q_{k,j}$ and $Q_{i,l}$ are the same, because $\forall n\in \mathbb{N}$, $5^n>4(5^{n-1}+\cdots +5+1)$.

It is easy to see that $Q_{k,j}(5)$ is divisible by $5$. Moreover, $Q_{k,j}(5)\ne 0$ (otherwise $Q_{k,j}$ is the zero polynomial which is impossible) and $|Q_{k,j}(5)|\le |P_i(5)|+|P_j(5)|\le n^2$. Therefore, $Q_{k,j}(5)$ can take at most $\frac{2}{5}{n}^2$ different values. But the overall number of polynomials $Q_{k,j}$ is bigger than $\frac{2}{5}{n}^2$, because $$\frac{1}{2}n(n-2)>\frac{2}{5}{n}^2\iff 5(n-2)>4n\iff n>10.$$ This implies that $Q_{k,j}(5)=Q_{i,l}(5)\Rightarrow Q_{k,j}=Q_{i,l}$ for some $i,j,k,l$. Consequently, $P_i+P_j=P_k+P_l$ for some $1\le i,j,k,l\le n$, $\{i,j\}\ne \{k,l\}$ which finishes the proof.