Find the least possible value of $n, n \geq 2015$ such that there exists polynomial $P(x)$ with degree $n$, integer coefficients, the coefficient of the term $x^n$ is positive and polynomial $Q(x)$ with integer coefficients satisfying the following equation for all $x \in \mathbb{R}$: $$x(P(x))^2 - 2P(x) = (x^3-x)(Q(x))^2$$

Source: This problem is from my school competition. I tried a few attempts in the exam room but they didn't seem to work. So far I concluded $degQ=degP-1$ and the two leading coefficients of $P(x)$ and $Q(x)$ are equal (assuming that the first coefficient of $Q(x)$ is positive, otherwise we could consider $-Q(x)$ instead), and $x$ divides $P(x)$


Let us first consider the equation$$(U(x))^2 - 1 = (x^2 - 1)(V(x))^2,\tag*{$(1)$}$$with $U$, $V \in \mathbb{R}[x]$, with highest coefficients of $U(x)$, $V(x)$ both positive.

Lemma 1. $(1)$ has at most one solution per degree of $P(x) \ge 1$.

Proof. Let $k$ be the degree of $U(x)$. We get $n > 0$, else we would have $V(x) = 0$ for all $x$, whose highest coefficient is not positive. $U(x)$ and $V(x)$ have no common root. Taking the derivative, we get then that $V(x)$ divides $U'(x)$ and so, since they have same degrees, $V(x) = \alpha U'(x)$. The equation is then$$(U(x))^2 - 1 = \alpha^2(x^2 - 1)(U'(x))^2.$$Taking the derivative, we get$$U(x) = \alpha^2(xU'(x) + (x^2 - 1)U^{\prime\prime}(x)).$$Looking at coefficients identification, we get that $\alpha^2 = 1/k^2$ and that all coefficients are fully defined when $a_n$ is chosen. Then plugging $x = 1$ into the original equation, we get $U(1)^2 = 1$ and so $a_k$ is determined and so at most one solution per degree.$$\tag*{$\square$}$$

Lemma 2. $(1)$ has exactly one solution per degree of $U(x) \ge 1$.

Proof. Let $U_k(x)$ be the uniquely polynomial defined as $\cos kx = U_k(\cos x)$. Let $V_k(x)$ be the unique polynomial defined as $\sin kx = (\sin x)V_k(\cos x)$. The degree of $U_k(x)$ is $k$ while the degree of $V_k(x)$ is $k - 1$. The highest coefficient of $U_k(x)$ is greater than $0$. We have$$(U_k(\cos x))^2 - 1 = \cos^2 kx - 1 = -\sin^2 kx = -(\sin^2x) (V_k(\cos x))^2 = (\cos^2x - 1)(V_k(\cos(x))^2.$$And so$$(U_k(x))^2 - 1 = (x^2 - 1)(V_k(x))^2.$$And so, for degree $k \ge 1$, the only solution to $(1)$ is $(U_k(x), \pm V_k(x))$, where the $\pm$ depends on the sign of the highest degree coefficient in $V_k(x)$.$$\tag*{$\square$}$$We head back to the original equation. The equation may be written as$$(xP(x) - 1)^2 - 1 = (x^2 - 1)(xQ(x))^2.$$And so $xP(x) - 1 = U_k(x)$ since highest coefficients are positive, and $xQ(x) = \pm V_k(x)$.

  • If $n \equiv 1 \text{ mod }2$: $U_k(0) = U_k(\cos \pi/2) = \cos k\pi/2 = 0$, and so no suitable $P(x)$.
  • If $n \equiv 0 \text{ mod }4$: $U_k(0) = U_k(\cos \pi/2) = \cos k\pi/2 = 1$, and so no suitable $P(x)$.
  • If $n \equiv 2 \text{ mod }4$: $U_k(0) = U_k(\cos \pi/2) = \cos k\pi/2 = -1$, and one suitable $P(x) = (U_k(x) + 1)/x$.

And so solutions exist only when the degree of $P(x)$ is in the form $4n + 1$. Hence, the answer is $n = 2017$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.