Floor function for polynomials

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Floor function for polynomials

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Source: 2025 Utah Math Olympiad #2

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kred9

#1 h Apr 5, 2025, 11:53 PM • 1 Y

Y by PikaPika999

Given polynomials $f(x)$ and $g(x)$ , where $g(x)$ is not the zero polynomial, we define $\left \lfloor \frac{f(x)}{g(x)} \right \rfloor$ to be the unique polynomial $q(x)$ such that we can write $f(x)=g(x)\cdot q(x) + r(x)$ , where $r(x)$ is a polynomial such that either $r(x)=0$ or the degree of $r(x)$ is less than the degree of $g(x)$ . Find all polynomials $p(x)$ with real coefficients such that $\left \lfloor \frac{p(x)}{x} \right \rfloor + \left \lfloor \frac{p(x)}{x+1} \right \rfloor =x^2.$

This post has been edited 1 time. Last edited by kred9, Apr 5, 2025, 11:53 PM
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ND_

#2 h Apr 6, 2025, 2:50 AM

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Since degree of $p(x)$ is 3, let $p(x)=ax^3+bx^2+cx+d$ . Then,
$\left \lfloor \frac{p(x)}{x} \right \rfloor=ax^2+bx+c,$ $\left \lfloor \frac{p(x)}{x+1} \right \rfloor = ax^2 - 3ax + bx + 3a-2b+c$
Solving, we get $(a,b,c,d)=\frac{1}{2}, \frac{3}{4}, 0, d$ , so the answer is $\boxed{\frac{1}{2}x^3+\frac{3}{4}x^2+d}$

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KAME06

#3 h Apr 6, 2025, 3:44 AM • 2 Y

Y by kred9, teomihai

By the definition, let $\left \lfloor \frac{p(x)}{x} \right \rfloor= q_1(x)$ and $\left \lfloor \frac{p(x)}{x+1} \right \rfloor= q_2(x)$ . Then:
$p(x)=x \cdot q_1(x)+C_1$ and $p(x)=(x+1) \cdot q_2(x)+C_2$ , where $C_1, C_2 \in \mathbb{R}$ . Let $n$ the degree of $P(x)$ . We deduce $q_1(x), q_2(x)$ are degree $n-1$ .
The problem tell us that $q_1(x) + q_2(x)=x^2$ , so $n-1 \ge 2 \Rightarrow n \ge 3$ .
Also, adding: $2p(x)=x \cdot q_1(x)+C_1+(x+1) \cdot q_2(x)+C_2=x(q_1(x)+q_2(x))+q_2(x)+C_1+C_2=x^3+q_2(x)+C_1+C_2$ .
$LHS$ is degree $n$ and, if $n-1 \ge 3$ , $RHS$ is degree $n-1$ . Contradiction. So $n-1 < 3 \Rightarrow n \le 3$ .
We conclude that $n=3$ . Then $q_1(x) + q_2(x)=x^2 \Rightarrow q_1(x)=ax^2+bx+d$ and $q_2(x)=(1-a)x^2-bx-d$ .
$p(x)=p(x) \Rightarrow x \cdot (ax^2+bx+d)+C_1 = (x+1) \cdot ((1-a)x^2-bx-d)+C_2$
$\Rightarrow ax^3+bx^2+dx+C_1 = (1-a)x^3+(1-a-b)x^2+(-d-b)x-d+C_2$
$\Rightarrow \begin{cases} a=1-a \\ b=1-a-b \\ d=-d-b \\ C_1=d+C_2 \end{cases}$
Solving the system: $(a, b, d, C_1)=(\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, C)$ , where $C \in \mathbb{R}$ . Then $p(x)=\frac{1}{2}x^3+\frac{1}{4}x^2 -\frac{1}{8}x+ C$ .
Checking:
$p(x)=x\left(\frac{1}{2}x^2+\frac{1}{4}x -\frac{1}{8}\right)+ C$
$p(x)=(x+1)\left(\frac{1}{2}x^2-\frac{1}{4}x +\frac{1}{8}\right)+C-\frac{1}{8}$
Then $\left \lfloor \frac{p(x)}{x} \right \rfloor + \left \lfloor \frac{p(x)}{x+1} \right \rfloor =\frac{1}{2}x^2+\frac{1}{4}x -\frac{1}{8}+\frac{1}{2}x^2-\frac{1}{4}x +\frac{1}{8}=x^2$

This post has been edited 2 times. Last edited by KAME06, Apr 6, 2025, 4:30 AM

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