Problem 10-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not intersect pairwise?

Answer: 2.
Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+q$ or $x^{2}+x+q$.

The ranges of any two polynomials of different forms intersect: indeed, the values of the polynomials $x^{2}+q$ and $x^{2}+x+q^{\prime}$ coincide at $x=q-q^{\prime}$. Therefore, polynomials of different forms cannot be chosen.

Polynomials of the first form can be chosen no more than two, since if the ranges of $f_{1}(x)=x^{2}+q$ and $f_{2}(x)=x^{2}+q^{\prime}$ do not intersect, then $q-q^{\prime}=4 k+2$ for some $k \in \mathbb{Z}$. Indeed, for an odd difference of the constant terms $q-q^{\prime}=2 k+1$, we have $f_{1}(k)=f_{2}(k+1)$. For a difference of the constant terms divisible by 4, $q-q^{\prime}=4 k$, we have $f_{1}(k-1)=f_{2}(k+1)$. But if at least three polynomials are chosen, then among the pairwise differences of the constant terms, at least one does not have the form $4 k+2$.

Polynomials of the second form can also be chosen no more than two, since if the ranges of $f_{1}(x)=x^{2}+x+q$ and $f_{2}(x)=x^{2}+x+q^{\prime}$ do not intersect, then $q-q^{\prime}=2 k+1$ for some $k \in \mathbb{Z}$. Indeed, for an even difference of the constant terms $q-q^{\prime}=2 k$, we have $f_{1}(k-1)=f_{2}(k)$. Again, if at least three polynomials are chosen, then among the pairwise differences of the constant terms, at least one is even.

Thus, more than two polynomials cannot be chosen. Example for two: $f_{1}(x)=x^{2}$ and $f_{2}(x)=x^{2}+2$.

| Grade | Score | Content criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Solution is correct modulo minor inaccuracies |
| $+/-$ | 14 | There is a proof that for each type of values $\left(x^{2}+q\right.$ and $\left.x^{2}+x+q\right)$ no more than two trinomials can be taken |
| $-/+$ | 8 | Complete solution, but only for the case $x^{2}+q$ |
| .- | 4 | Example of two trinomials whose ranges do not intersect |
| - | 0 | Solution is completely incorrect/only answer |