2016.3.3 Question 3

1. We have that

   $$\begin{array}{llll}\hfill \frac{d}{dx}\frac{e^{x}P(x)}{Q(x)}& =\frac{Q(x)\left[e^x{P}^{\prime }(x)+e^{x}P(x)\right]-Q^{\prime }(x)e^{x}P(x)}{Q{(x)}^2}& \hfill & \\ \hfill & =e^x\frac{\left[Q(x)P^{\prime }(x)+Q(x)P(x)-Q^{\prime }(x)P(x)\right]}{Q{(x)}^2}& \hfill & \\ \hfill & =e^x\frac{x^3-2}{{(x+1)}^2}.& \hfill & \end{array}$$

   Therefore, we have

   $$\begin{array}{llll}\hfill \frac{\left[Q(x)P^{\prime }(x)+Q(x)P(x)-Q^{\prime }(x)P(x)\right]}{Q{(x)}^2}& =\frac{x^3-2}{{(x+1)}^2}& \hfill & \\ \hfill {(x+1)}^2\left[Q(x)P^{\prime }(x)+Q(x)P(x)-Q^{\prime }(x)P(x)\right]& =Q{(x)}^2\left(x^3-2\right).& \hfill & \end{array}$$

   If we plug in $x=-1$ on both sides, we have $\text{LHS}=0$ and $\text{RHS}=Q{(-1)}^2\cdot (-3)$.

   Therefore, $Q{(-1)}^2=0$, $Q(-1)=0$.

   Since $Q(x)\in \mathbb{P}[x]$, we must have

   $$(x+1)\mid Q(x)$$

   as desired.

   Therefore, $\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q\ge 1$, $\text{deg}<!--\; FUNCTION\; APPLICATION\; -->\text{RHS}=3+2\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q$.

   If $\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P=-\infty$, $P(x)=0$,$\text{LHS}=0$ which is impossible.

   If $\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P=0$, $P(x)=C\in ℝ\setminus \{0\}$, $\text{LHS}=C{(x+1)}^{2}Q(x)$, $\text{deg}<!--\; FUNCTION\; APPLICATION\; -->\text{LHS}=\text{deg}<!--\; FUNCTION\; APPLICATION\; -->q+2$, which is impossible.

   Therefore, we have $\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P^{\prime }=\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P-1$. Hence,

   $$\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q(x)P^{\prime }(x)=\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P^{\prime }(x)Q(x)=\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P+\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q-1,$$

   and

   $$\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q(x)P(x)=\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P+\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q.$$

   Therefore,

   $$\text{deg}<!--\; FUNCTION\; APPLICATION\; -->\text{LHS}=2+\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P+\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q=\text{deg}<!--\; FUNCTION\; APPLICATION\; -->\text{RHS},$$

   which gives

   $$\text{deg}<!--\; FUNCTION\; APPLICATION\; -->P=\text{deg}<!--\; FUNCTION\; APPLICATION\; -->Q+1,$$

   as desired.

   When $Q(x)=x+1$, let $P(x)=a{x}^2+\mathit{bx}+c$ where $a\ne 0$. We have $P^{\prime }(x)=2\mathit{ax}+b$. Therefore,

   $$\begin{array}{llll}\hfill {(x+1)}^2\left[Q(x)P^{\prime }(x)+Q(x)P(x)-Q^{\prime }(x)P(x)\right]& =Q{(x)}^2\left(x^3-2\right)& \hfill & \\ \hfill Q(x)P^{\prime }(x)+Q(x)P(x)-Q^{\prime }(x)P(x)& =x^3-2& \hfill & \\ \hfill (x+1)(2\mathit{ax}+b)+(x+1)(a{x}^2+\mathit{bx}+c)-(a{x}^2+\mathit{bx}+c)& =x^3-2& \hfill & \\ \hfill (x+1)(2\mathit{ax}+b)+x(a{x}^2+\mathit{bx}+c)& =x^3-2& \hfill & \\ \hfill a{x}^3+(2a+b)x^2+(2a+b+c)x+b& =x^3-2.& \hfill & \end{array}$$

   This solves to $(a,b,c)=(1,-2,0)$. Therefore, $P(x)=x^2-2x$.

2. In this case, we must have that

   $$(x+1)\left[Q(x)P^{\prime }(x)+Q(x)P(x)-Q^{\prime }(x)P(x)\right]=Q{(x)}^2.$$

   Therefore, $Q(x)=(x+1)R(x)$ for some $R(x)\in \mathbb{P}[x]$. We may assume $P(-1)\ne 0$.

   Hence, $Q^{\prime }(x)=(x+1)R^{\prime }(x)+R(x)$

   Plugging this in gives us

   $$(x+1)R(x)P^{\prime }(x)+(x+1)R(x)P(x)-\left[(x+1)R^{\prime }(x)+R(x)\right]P(x)=(x+1)R{(x)}^2,$$

   which simplifies to

   $$(x+1)\left[R(x)P^{\prime }(x)+R(x)P(x)-R^{\prime }(x)P(x)\right]-R(x)P(x)=(x+1)R{(x)}^2.$$

   Let $x=-1$, and we can see $x+1$ divides $R(x)$, since $x+1$ can’t divide $P(x)$.

   Therefore, let $R(x)=(x+1)S(x)$, therefore $R^{\prime }(x)=S(x)+(x+1)S^{\prime }(x)$.

   This gives

   $$(x+1)S(x)\left[P^{\prime }(x)+P(x)\right]-\left[S(x)+(x+1)S^{\prime }(x)\right]P(x)-S(x)P(x)={(x+1)}^{2}S{(x)}^2,$$

   which simplifies to

   $$(x+1)\left[S(x)P^{\prime }(x)+S(x)P(x)-S^{\prime }(x)P(x)\right]-2S(x)P(x)={(x+1)}^{2}S{(x)}^2.$$

   Therefore, we can see that $x+1$ divides $S(x)$ by similar reasons.

   Repeating this, we can conclude that there are arbitrarily many factors of $x+1$ in $Q(x)$ (proof by infinite descent), which is impossible.

   Formally speaking, let $Q(x)={(x+1)}^{n}T(x)$ where $T(-1)\ne 0$, $n\in \mathbb{N}$. Therefore, we have

   $$\begin{array}{llll}\hfill Q^{\prime }(x)& =n{(x+1)}^{n-1}T(x)+{(x+1)}^n{T}^{\prime }(x)& \hfill & \\ \hfill & ={(x+1)}^{n-1}\left[\mathit{nT}(x)+(x+1)T^{\prime }(x)\right].& \hfill & \end{array}$$

   Therefore,

   $$(x+1)\left[Q(x)P^{\prime }(x)+Q(x)P(x)-Q^{\prime }(x)P(x)\right]=Q{(x)}^2$$

   simplifies to

   $${(x+1)}^{n+1}T(x)\left[P^{\prime }(x)+P(x)\right]-{(x+1)}^n\left[\mathit{nT}(x)+(x+1)T^{\prime }(x)\right]P(x)={(x+1)}^{2n}T{(x)}^2,$$

   which further simplifies to

   $$(x+1)\left[T(x)P^{\prime }(x)+T(x)P(x)-T^{\prime }(x)P(x)\right]-\mathit{nT}(x)P(x)={(x+1)}^{n}T{(x)}^2.$$

   Now, let $x=-1$, we have that $\mathit{nT}(-1)P(-1)=0$. But $n\ne 0$, $T(-1)\ne 0$, $P(-1)\ne 0$, which gives a contradiction.

   Therefore, such $P$ and $Q$ do not exist.