1. Do there exist natural numbers $a, b$, and $c$ such that each of the equations

$$
\begin{array}{ll}
a x^{2}+b x+c=0, & a x^{2}+b x-c=0 \\
a x^{2}-b x+c=0, & a x^{2}-b x-c=0
\end{array}
$$

has both roots as integers?
1. The solution can simply consist of checking that the numbers $a=1, b=5, c=6$ indeed satisfy the condition. Let's try to explain how such numbers can be found. First, let's try to simplify our task. Notice, first of all, that the roots of equations differing only in the sign of $b x$ are opposite, so it is sufficient to ensure that the roots of the equations $a x^{2}+b x+c=0$ and $a x^{2}+b x-c=0$ are integers. Then the roots of the equations $a x^{2}-b x+c=0$ and $a x^{2}-b x-c=0$ will automatically be integers.

Furthermore, if $x_{1}$ and $x_{2}$ are integer roots of the equation $a x^{2} + b x + c$, then by Vieta's theorem $b=-\left(x_{1}+x_{2}\right) a, c=x_{1} x_{2} a$, i.e., all coefficients of the equation can be reduced by $a$, and again we will get an equation with natural coefficients and the same roots. Therefore, without loss of generality, we can assume that $a=1$.

If $x_{1}$ and $x_{2}$ are the roots of the first equation, then by Vieta's theorem $x_{1} x_{2}=c, x_{1}+x_{2}=-b$. Similarly, if $y_{1}$ and $y_{2}$ are the roots of the second equation, then by Vieta's theorem $y_{1} y_{2}=-c$, $y_{1}+y_{2}=-b$. It remains to find such $x_{1}, x_{2}, y_{1}, y_{2}$ that

$$
x_{1}+x_{2}=y_{1}+y_{2}, \quad x_{1} x_{2}=-y_{1} y_{2}
$$

For example, $x_{1}=-2, x_{2}=-3, y_{1}=1, y_{2}=-6$ work.

In the arithmetic solution. For the roots of the equations $x^{2}+b x+c=0$ and $x^{2}+b x-c=0$ to be integers, it is necessary that the discriminants of these equations be perfect squares: $b^{2}-4 c=m^{2}, b^{2}+4 c=n^{2}$. This condition turns out to be sufficient as well - since $b^{2}-4 c$ has the same parity as $b$, and thus $\sqrt{b^{2}-4 c}$ (if this root is an integer) has the same parity (see fact 23). Thus, $x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 c}}{2}$ (the roots of the equation $x^{2}+b x+c=0$) are integers. A similar argument is valid for the second equation.

In fact, there are infinitely many such $b, c, m, n$ such that $b^{2}-4 c=m^{2}$, $b^{2}+4 c=n^{2}$. To solve the problem, it is sufficient to provide one example of such $b, c$, $m$, $n$. Let's show how such an example could be found. By eliminating $c$, we get $n^{2}-b^{2}=b^{2}-m^{2}$, and to ensure that $c$ is an integer, it is necessary that $b, m$, and $n$ have the same parity. By trying the squares of odd numbers from 1 to 9, we find one solution: $7^{2}-5^{2}=5^{2}-1=24$; $c=24 / 4=6$. We have found $b$ and $c$ such that the discriminants of both equations are perfect squares.

Comment. To find an example, we needed to find a representation of $2 b^{2}$ as a sum $m^{2}+n^{2}$, different from $b^{2}+b^{2}$. We found such $b, m$, and $n$ by trial and error, but in fact, the problem of representing a given number as a sum of two squares is well studied. See the comment to problem 4 for 11th grade of the 1996 competition.