Number of Polynomials with Integer Coefficients that are bounded by $x^2$ and $x^4+1$ What is the number of polynomials $p(x)$ with integer coefficients, such that
$x^2≤p(x)≤x^4+1$ for all real numbers $x$?
 A: Hint : No polynomial with degree $3$ can have this property because it has negative values. It is easy to find out the polynomials with degree $2$ and $4$ satisfying the given bounds. Note that $x^2>x^4$ for $0<x<1$
A: Indeed, no polynomial with degree $3$ can satisfy this property because it has to have a negative value at some point.
Also, it is clear that (by taking very large $x$), the degree of this polynomial must be $2,3$ or $4$.
So this polynomial has to have a degree of $2$ or $4$.
Case 1. deg$p$ is $2$.
Let $p(x)=ax^2+bx+c$. Since $0 \le p(0)=c \le 1$, we either have $c=0$ or $c=1$.
If $c=0$, we have $p(x)=ax^2+bx$. For $p(x) \ge x^2$ to hold for all $x$, we must have $a \ge 1$. 
Also, the determinant of $p(x)-x^2$ must be $\le 0$ for all $x$. 
The determinant is just $b^2$, so $b=0$. Now $p(x)=ax^2$.
Since $x^4+1 \ge ax^2$, we take $x=1$ so that $a \le 2$. AM-GM gives $x^4+1 \ge 2x^2 \ge x^2$, so $a=1,2$ works. 
The solution set for this case is $p(x)=x^2$ and $p(x)=2x^2$.
Case 2. deg$p$ is $4$.
Denote the coefficient of $x^4$ as $l$, and the constant term as $e$.
Clearly, $0 \le l \le 1$ by taking a large $x$. Since $l$ is nonzero, $l=1$.
Again, we have $0 \le e \le 1$ by taking $x=0$. 
The coefficient of $x^3$ must be $0$. Otherwise, $x^4+1-p(x)$ can take negative values as it is a degree $3$ polynomial.
Let $p(x)=x^4+ax^2+bx+e$.
SubCase 2-1. $e=1$.
We have $-ax^2-bx \ge 0$ for all $x$. We must have $a \le 0$. Since the determinant of $-ax^2-bx$ is less than $0$, we have $b^2 \le 0$, so $b=0$.
We now have $x^4+ax^2+1 \ge x^2$. Take $x=1$ to get $a \ge -1$. 
AM-GM gives $x^4+1 \ge x^4-x^2+1 \ge x^2$, so both $a=0$ and $a=-1$ work.
The solution set for this subcase is $p(x)=x^4+1$ and $x^4-x^2+1$.
SubCase 2-2. $e=0$.
We have $-ax^2-bx+1 \ge 0$. We must have $a \le 0$, and the determinant $b^2+4a \le 0$. 
Also, setting $x=1$ gives $a+b \ge 0$. 
Now we have $b^2 \le -4a \le 4b$, giving $b(b-4) \le 0$, so $b=0,1,2,3,4$.
If $b=0$, we have $a \ge 0$ and $a \le 0$, so $a=b=0$. This gives $p(x)=x^4$, which fails.
If $b=1$, we have $4a \ge -1$, so $-\frac{1}{4} \le a \le 0$. This gives $a=0$.
This gives $p(x)=x^4+x$, which fails at $x=2$.
If $b=2$, we have $4a \ge -4$, so $-1 \le a \le 0$.
If $a=0$, we have $p(x)=x^4+2x$, which fails at $x=1$.
If $a=-1$, we have $p(x)=x^4-x^2+2x$, which fails at $x=-1$.
If $b=3$, we have $4a \ge -9$, so $-\frac{9}{4} \le a \le 0$, giving $a=-2,-1,0$.
If $a=0$, we have $p(x)=x^4+3x$, which fails at $x=1$.
If $-2 \le a \le -1$, we have $p(x)=x^4+ax^2+3x$, which fails at $x=-1$.
If $b=4$, we have $-4 \le a \le 0$.
If $a=0$, we have $p(x)=x^4+4x$, which fails at $x=1$.
If $-4 \le a \le -1$, we have $p(x)=x^4+ax^2+4x$, which fails at $x=-1$.
Therefore, there are no solutions in this subcase.
The final answer is $x^2$, $2x^2$, $x^4+1$, $x^4-x^2+1$.
Hopefully, someone can find a cleaner solution..
