Do there exist polynomials $p(x)$ and $q(x)$ with real coefficients such that $p^3(x)-q^2(x)$ is linear but not constant?
Problem
Source: Kürschák 2017 Problem 2
Tags: Polynomials, algebra, polynomial
07.10.2017 15:12
Not sure if this works but I'll post it anyway so let's suppose you divide by the appropriate constants for $p$ and $q$ so that $p(x)^3-q(x)^2=\pm x-\alpha$ for a real number $\alpha$. Notice that there is an integer $n$ with $\deg(p)=2n$ and $\deg(q)=3n$. Then differentiate to get \[3p(x)^2p'(x)-2q(x)q'(x)=\pm 1\]and so using these two equations you can figure out \[q(x)\mid p(x)^3\mp x+\alpha\]\[q(x)\mid 3p(x)^2p'(x)\mp 1.\]Now notice that these imply $q(x)\mid 3p^3(x)p'(x)\mp p(x)$ and so $q(x)\mid 3p'(x)(\pm x-\alpha)\mp p(x)$, but then $\deg(q)>\deg(RHS)$ and so it must be that $3p'(x)(\pm x-\alpha)\mp p(x)=0$, or $\pm x-\alpha\mid p(x)$. Then $\pm x-\alpha\mid g(x)$ as well, and you can get an easy contradiction from here (as $(\pm x-\alpha)^2\mid LHS$ but not $RHS$ in original equation). @below (rafayaaaaaashary1) thanks for pointing out relevant things
07.10.2017 15:19
$p(x)=x^1/3 and q(x)=\sqrt(x+\alpha)$
07.10.2017 15:22
Giffunk wrote: $p(x)=x^1/3 and q(x)=(x+\alpha)^1/2$? Then the difference will be a degree $3$ polynomial.
07.10.2017 15:24
Giffunk wrote: $p(x)=x^{1/3}\text{ and }q(x)=(x+\alpha)^{1/2}$? These are not polynomials Hydrogen-Helium wrote: Not sure if this works but I'll post it anyway so let's suppose you divide by the appropriate constants for $p$ and $q$ so that $p(x)^3-q(x)^2=x-\alpha$ for a real number $\alpha$. I don't see anything overtly wrong, but you must be cautious in this line; $i\not\in \mathbb R$ so we cant divide $q(x)^2$ by $-1$. Fortunately this is a highly nonfatal error, seeing as we can just replace $x-\alpha$ with $\pm(x-\alpha)$
07.10.2017 15:25
Yeah saw this now sorry about that
11.10.2017 15:32
<jatekos101 wrote: Do there exist polynomials $p(x)$ and $q(x)$ with real coefficients such that $p^3(x)-q^2(x)$ is linear but not constant? The following is just simplification of the solution by HHe. I post it for completeness and simplicity. Suppose that there exists such polynomials, say $p(x)^3 -q(x)^2 = ax+b$ where $a \neq 0$. Since one may substitute $t \rightarrow ax+b$ and then $x \rightarrow t$, we may assume $p(x)^3-q(x)^2 = x$. Then $p^3 \equiv x \pmod{q^2}$. Thus $3p^2p' \equiv 1 \pmod{q}$. Multiplying $p$ on both sides, we have $3xp' \equiv p \pmod{q}$, i.e. $q| p-3xp'$. Note that degree of $q$ exceeds that of $p$. Thus $p=3xp'$. From this now, there are many ways to leads to an absurdity. One is to observe that $x|p$. It follows that $x|q$. But then $x^2|p^3-q^2$ whereas $x^2 \nmid x$. One other way is to observe that $p=3xp'$ implies degree of $p$ is $\frac{1}{3}$ which is impossible.
29.10.2017 13:37
My solution also works using derivatives.
11.06.2024 20:48
Cute! We claim that there does not exist such $p,q$ Assume $p(x)^3-q(x)^2 =r(x)$ for linear $r$. Claim 1: $p(x),q(x)$ are co-prime. Proof: Assume $(p(x),q(x))=s(x)$ so that $\deg (s(x)) \geq 1$ then $s(x)^2 \mid r(x)$ contradiction as $\deg(s(x)^2) \geq 2 \geq \deg(r(x))$ $\blacksquare$ As this is an polynomial identity, taking derivatives both sides yields: $$3p(x)^2p'(x)-2q(x)q'(x) = r'(x)$$and therefore $$(p(x)^3-q(x)^2)r'(x)=(3p(x)^2p'(x)-2q(x)q'(x))r(x)$$$$\therefore p(x)^2 \mid qr'(x)-2q'(x)r(x)$$however this is a contradiction as $2\deg(p(x)) \geq \deg(qr'(x)-2q'(x)r(x)) = \frac{3 \deg(p(x))}{2}$.