Problem 5. (3 points)
$P(x)$ is a polynomial of the fourth degree with integer coefficients, the leading one of which is positive. Moreover, $P(\sqrt{3})=P(\sqrt{5})$. Find the value(s) of $x$ for which $P(x)$ takes its minimum value.
# Answer: $-2 ; 2$

## Solution:

Let $P(x)=a x^{4}+b x^{3}+c x^{2}+d x+e$. Then $P(\sqrt{3})=9 a+3 \sqrt{3} b+3 c+\sqrt{3} d+e, P(\sqrt{5})=$ $25 a+5 \sqrt{5} b+5 c+\sqrt{5} d+e$. Numbers of the form $A+B \sqrt{3}$ and $C+D \sqrt{5}$ with integer $A, B, C, D$ can only be equal if $B=D=0$.

Thus, $3 b+d=5 b+d=0$, from which $b=d=0$, meaning $P(x)=a x^{4}+c x^{2}+e$ and we can say that $P(x)=Q\left(x^{2}\right)$, where $Q(t)=a t^{2}+c t+d$. In this case, $Q(3)=Q(5)$, so $Q(t)$ reaches its minimum at $t=\frac{3+5}{2}=4$, and $P(x)$ reaches its minimum at $x= \pm \sqrt{4}= \pm 2$.