# Problem 1. (2 points)

Quadratic equations $x^{2}+p x+q$ and $x^{2}+a x+b$ have one root each. Among the numbers $p, q, a, b$ are 16, 64, and 1024. What can the fourth number be?

If there are multiple possible answers, enter the larger one in the system, and indicate all in the written solution.
Answer: $\pm 8, \pm 64,262144$

## Solution:

The discriminant of the equation $x^{2}+p x+q$ must be equal to zero, so $p^{2}=4 q$. Similarly, $a^{2}=4 b$.

Therefore, if $p$ is 16, 64, or 1024, then $q=\frac{p^{2}}{4}$, which means 64, 1024, or $2^{18}=262144$. Similarly for $a$ and $b$.

If $q$ is 16, 64, or 1024, then $p= \pm 2 \sqrt{q}$, which means $\pm 8, \pm 16$, or $\pm 64$. Similarly for $a$ and $b$.

Therefore, the coefficient 64 must be in the same equation with 16 or 1024, and the fourth number is $\pm 8, \pm 64$, or 262144.