4. Teacher Maria Ivanovna is preparing tasks for a math lesson. She wants to substitute three different natural numbers for $a$, $b$, and $c$ in the equation $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ so that the roots of the equation are integers. Help her: choose such numbers and solve the equation.
Solution. There are many such equations, it is sufficient to provide one of them. For example, the roots of the equation

$$
\frac{1}{x+6}+\frac{1}{x+3}=\frac{1}{2}
$$

are the numbers 0 and -5.

Comment. The equation provided in the solution can be derived. Consider the equality

$$
\frac{1}{6}+\frac{1}{3}=\frac{1}{2}
$$

We can immediately take $c=2$ and make one of the roots equal to 0. This gives us the equation mentioned above. It remains to verify that the second root is also an integer.

From the reasoning provided below, it follows that the second root must necessarily be an integer. Indeed, solving the given equation reduces to solving a quadratic equation with integer coefficients. By Vieta's theorem, if one root is an integer, then the other must also be an integer.

How can we conduct a targeted search for such numbers $a, b$, and $c$? The given equation can be reduced to the quadratic equation

$$
x^{2}+(a+b-2 c) \cdot x+(a b-a c-b c)=0
$$

Its discriminant is $(a-b)^{2}+4 c^{2}$. We need to choose such numbers so that the discriminant is equal to the square of an integer. This can be done by recalling "Pythagorean" triples, for example, $3^{2}+4^{2}=5^{2}$ or $5^{2}+12^{2}=13^{2}$.

In the first case, we can take $c=2, a-b=3$. Then

$$
x_{1,2}=\frac{2 c-a-b \pm 5}{2}=\frac{4-a-b \pm 5}{2}
$$

For $a=6, b=3$, we have $x_{1}=0, x_{2}=-5$. This gives us the equation mentioned above:

$$
\frac{1}{x+6}+\frac{1}{x+3}=\frac{1}{2}
$$

In the second case, similarly, we take $c=6, a=7, b=2$. We get the equation

$$
\frac{1}{x+7}+\frac{1}{x+2}=\frac{1}{6}
$$

the roots of which are the numbers -5 and 8.

## Grading Criteria.

- Full solution - an equation with specific values of $a, b$, and $c$, which is solved and its roots are found, - 7 points.
- The solution provides a suitable equation without solving it (or solved incorrectly), and the jury verifies that it fits, - 3 points.