Problems from Ukrainian Authors

Since the polynomials $P_1, \dots, P_n$ do not have integer roots then $P_i(0) \neq 0$ 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 $k \neq j$ 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 $\forall n \in \mathbb{N}$, $5^n > 4(5^{n-1}+\dots+5+1)$.

It is easy to see that $Q_{k,j}(5)$ is divisible by $5$. Moreover, $Q_{k,j}(5) \neq 0$ (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
$$\frac{1}{2}n(n-2) > \frac{2}{5}n^2 \Leftrightarrow 5(n-2) > 4n \Leftrightarrow n > 10.$$
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 $1 \le i,j,k,l \le n$, $\{i, j\} \neq \{k, l\}$ which finishes the proof.