# Problem 1.

When designing a power plant, it became necessary to solve the equation

$$
4 x^{4}+4 p x^{3}=(p-4) x^{2}-4 p x+p,
$$

where $p$ is an integer parameter specified by the developer. For quick and reliable solutions, it is required that the roots of the equation be rational numbers. For which $p$ is this true?
# Solution.

The problem boils down to finding the roots of the polynomial

$$
f(x, p)=\left(x^{2}+1\right)\left(4 x^{2}+4 p x-p\right) .
$$

For the discriminant of the quadratic trinomial, it is necessary that $D / 4=$ $4 p(p+1) \geq 0$.

If $p=0$ or $p=-1$, then there exist rational roots 0 and -1, respectively.

If $p>0$, then the roots are $x=-p / 2 \pm \sqrt{p(p+1)} / 2$.

We will show that the number $\sqrt{p(p+1)}$ is irrational. Assuming the opposite, we can represent it as an irreducible fraction $m / n$. Then $n^{2}\left(p^{2}+p\right)=m^{2}$, and consequently, $m$ is divisible by $n$, i.e., the fraction is reducible.

Answer: $p=0,-1$.