27. [16] Find the number of polynomials $p(x)$ with integer coefficients satisfying $p(x) \geq \min \left\{2 x^{4}-6 x^{2}+1,4-5 x^{2}\right\}$ and $p(x) \leq \max \left\{2 x^{4}-6 x^{2}+1,4-5 x^{2}\right\}$ for all $x \in \mathbb{R}$.
Answer: 4
We first find the intersection points of $f(x)=2 x^{4}-6 x^{2}+1$ and $g(x)=4-5 x^{2}$. If $2 x^{4}-6 x^{2}+1=4-5 x^{2}$, then $2 x^{4}-x^{2}-3=0$, so $\left(2 x^{2}-3\right)\left(x^{2}+1\right)=0$, and $x= \pm \sqrt{\frac{3}{2}}$. Note that this also demonstrates that $g(x) \geq f(x)$ if and only if $|x| \leq \sqrt{\frac{3}{2}}$ and that $p(x)$ must satisfy $p(x) \leq g(x)$ iff $|x| \leq \sqrt{\frac{3}{2}}$. We must have $p\left(\sqrt{\frac{3}{2}}\right)=-\frac{7}{2}$, so $p(x)=-\frac{7}{2}+\left(2 x^{2}-3\right) q(x)$ for a polnomial $q$ of degree 0,1 or 2 . We now examine cases.
Case 1: $q$ is constant. We have $q=\frac{n}{2}$ for an integer $n$ since $1 \leq p(0) \leq 4, n=-\frac{3}{2}$ or $-\frac{5}{2}$. Clearly $q=-\frac{5}{2}$ is an appropriate choice because then $p=g$. Let $p_{1}(x)=-\frac{7}{2}+-\frac{3}{2}\left(2 x^{2}-3\right)=1-3 x^{2}$. We have $p_{1}(x)$ is $\leq g(x)$ and $\leq f(x)$ in the right places, so this function works. Thus, there are 2 solutions. Case 2: $q$ is linear. We have $p(x)$ is a cubic, so $p(x)-g(x)$ is also a cubic, which means it can't be positive for both arbitrarily large positive values of $x$ and arbitrarily large negative values of $x$. Thus, there are no solutions.
Case 3: $q$ is quadratic. We have $q(x)=a x^{2}+b x+c$. Apply the same argument for the case when $q$ is a constant, we have $c=-\frac{3}{2}$ or $-\frac{5}{2}$. Since $p$ must have integer coefficients, we have $b$ must be an integer. Since $f(x) \geq p(x) \geq g(x)$ for large $x$, the leading coefficient of $f$ must be greater than or equal to the leading coefficient of $p$, which must be greater than 0 . Thus $a=1$ or $a=\frac{1}{2}$. However, if $a=\frac{1}{2}$, then the quadratic term of $p$ is not an integer, so $a=1$.
Now if $c=-\frac{5}{2}$, then $p(0)=4=g(0)$. But this is the maximum value of $g(x)$, so it must be a local maximum of $p$. Thus $p$ must not have a linear term odd (otherwise the function behaves like $-3 b x+4$ around $x=0$ ). So $p(x)$ must be $2 x^{4}-8 x^{2}+4$. This is indeed bounded between $f(x)$ and $g(x)$ at all points.
Now suppose $c=-\frac{3}{2}$. We have $a$ must equal to 1 . If $b \neq 0$, then $p$ will have a cubic term, which means $f(x)-p(x)$ can't be positive for both arbitrarily large positive $x$ and arbitrarily large negative $x$, so $b=0$. Therefore $p(x)=-\frac{7}{2}+\left(2 x^{2}-3\right)\left(x^{2}-\frac{3}{2}\right)=f(x)$. It is easy to check that this choice of $p$ is indeed bounded between $f(x)$ and $g(x)$.
Therefore, there are 4 solutions in total.