Р Рубанов и.с.

Znayka writes 10 numbers on the board, then Neznayka adds another 10 numbers, with all 20 numbers being positive and distinct. Could Znayka have written such numbers to ensure that he could later form 10 quadratic trinomials of the form $x^{2}+p x+q$, where the coefficients $p$ and $q$ include all the written numbers, and the (real) roots of these trinomials take exactly 11 different values?
Lemma 1. 1) If $p>4$ and $p>q$, then the quadratic polynomial $x^{2}+p x+q$ has two distinct roots.

2) If $00$, then at least one of the polynomials $x^{2}+p x+q, x^{2}+q x+p$ has no roots.

Proof. 1) The discriminant $p^{2}-4 q>4 q-4 q=0$.

2) If $q>p$, then the discriminant of the first polynomial $p^{2}-4 qr$ $=-\left(x_{0}+x_{2}\right)$, then $x_{1}x_{0} x_{2}=s$. Contradiction.

Second proof. The resultant of these polynomials is obviously positive (see problem 60925).

We will show that it is enough for Znayka to write the numbers $1 / 4,1 / 2,1,2,5,5^{2}, 5^{4}, 5^{8}, 5^{16}, 5^{32}$.

Consider all Nезнайкин's numbers greater than 4. If their quantity is odd, add one more Nезнайкин's number to them. Call these numbers marked. Add to the marked numbers from the set $5,5^{2}, 5^{4}, 5^{8}, 5^{16}, 5^{32}$ so that the total number of marked numbers is 12. If the powers of five are not enough, add a few more Nезнайкин's numbers. From the unused powers of five, form polynomials of the form $x^{2}+p x+q$, where $p<q$; then their discriminants are negative.

Write down the 12 marked numbers in ascending order: $p_{1}<p_{2}<\ldots<p_{12}$ and form six polynomials from them: $x^{2}+p_{12} x+p_{1}, x^{2}+p_{11} x+p_{2}, \ldots$, $x^{2}+p_{7} x+p_{6}$. By construction, among the 12 marked numbers, there are at least six greater than 4, so by Lemma 1, all these six polynomials have two roots. By Lemma 2, all these 12 roots are distinct.

Form the polynomial $x^{2}+2 x+1$. If its only root -1 appears among the roots of the marked polynomials, then declare the corresponding marked polynomial bad. If not, declare one of the marked polynomials bad. Remove the bad polynomial, and from its two coefficients and the numbers $1 / 4$ and $1 / 2$, form (by Lemma 1) two polynomials without roots. Now the number of distinct roots of the already formed polynomials is exactly 11.

There may be a few unused Nезнайкин's numbers left. All of them are no more than 4, and only one can equal 4. By Lemma 1, from them, we can form polynomials that have no roots.

## Answer

Can.