Ankoganit wrote: Note that the leading coefficient of $p(x)$ is uniquely determined by the leading coefficient of $r(x)$; $q(x^2)$ can't affect it anyway. I don't think its true (or maybe I misunderstand your idea). Consider this $p(x)=(ax^2+x)$ and $q(x)=-a^3x^3$. We have $r(x)=3a^2x^5+3ax^4+x^3$ as an odd degree polynomial. So $q(x)$ can influence it?

@anantmudgal09 Thanks, I've tried to fix it. Hopefully it's OK now.

There is a slight abuse of notation when you say $p(X)=X^mq(X)$ since $q$ is already used Anyway, i think this is correct but not so sure about this part, can you elaborate? Quote: With the constant term of one factor fixed, we can get only finitely many possible factorisations of $g(x)/x^{3m}$, and each of them leads to at most finite solutions. So finitely many possible $p(x)$, and hence also $q(x)$.

Ah, right. Loks lke I.m ouut of mt m!nd todey, makeing sk manyy miskates...

There are way too many traps to fall into, the situation gets quite counter-intuitive. For example, $\text{deg} p$ may be even, and $p(x)-p(-x)$ hardly gives any information about $p(x)$ due to only odd degree terms. For example, Ankoganit's solution is wrong. In fact, I don't think there's an easy fix, even though I want it to work. Here's why. Given $m$, the coefficient of the smallest-degree nonzero term $g_\ell x^\ell$ in $g(x)$ does not determine the constant term of $Q(x)$, because, say, for $Q(x)=c_0+c_1x+x^3+\dots$, $Q(x)-Q(-x)=2c_1x+2x^3+\dots$, and so the equation only yields $(2\pm 1)c_0^2\cdot c_1=g_\ell$, which does not determine $c_0$ whatsoever. Quite the troll problem. Only the official solution seems to be correct. The official solution considers this rather like a Diophantine equation (which is natural, given $\mathbb{R}[x]$ is a UFD).

There is such a proof I believe. The solution however is quite tricky in my opinion and is rather simple minded. Let $p(x)^3+q(x^2)=r(x)$ then $(p(x)^3-p(-x)^3)=r(x)-r(-x)$. Now $p(x)^3-p(-x)^3=(p(x)-p(-x))(p(x)^2+p(x)p(-x)+p(-x)^2)$ and note that the second factor is strictly positive. Given that there are only finitely many ways to factor $r(x)-r(-x)$ in such a form (note that $r(x)$ has odd degree so the polynomial $r(x)-r(-x)$ is nonzero and there are only finitely many ways to distribute the roots and then choose the initial coefficient) then are only finitely many pairs of polynomials $(p(x)-p(-x))$ and $(p(x)^2+p(x)p(-x)+p(-x)^2)$. This can be shown to determine $p(x)$ up to replacing it with $p(-x)$. (Remember we are given $r(x)$) Regardless it follows that the number of p(x) is finite and thus the number of $p(x),q(x)$ is finite.

How do you choose the leading coefficient? The main crux is proving that you cannot have infinitely many possible choices for it. Maybe the issue can be seen more clearly if we write $p(x)=u(x^2)+v(x^2)x$ etc. Then one factor is just $2v(x^2)x$, while the other factor has an independent $u$ part in it as well, which may bear a different leading coefficient.

Let $\omega=e^{\frac{2\pi i}{3}}$ be a third root of unity. By plugging in $x$ and $-x$ and taking the difference we need to prove that the equation $$r(x)-r(-x)=p(x)^3-p(-x)^3=\left(p(x)-p(-x)\right)\left(p(x)-\omega p(-x)\right)\left(p(x)-\omega^2 p(-x)\right)$$holds for finitely many real polynomials $p(x)$. Since $r(x)-r(-x)$ and $p(x)-p(-x)$ are both of the form $xc(x^2),$ it's enough to show that there cannot be infinitely many real polynomials $p(x)$ which satisfy $$t(x)-t(-x)=\left(p(x)-\omega p(-x)\right)\left(p(x)-\omega^2 p(-x)\right)$$for a fixed $t\in \mathbb{R}[X].$ FTSOC there are infinitely many $p\in \mathbb{R}[X]$ which satisfy the above equation. As $t(x)$ is fixed, monic polynomial in LHS of the equation can be represented in finitely many ways as the product of $2$ non-constant monic polynomials. Hence, by infinite PHP there exist fixed monic $a(x), b(x)\in \mathbb{C}[X]$ for which there are infinitely many real polynomials $p(x)$ which satisfy $$ua(x)=p(x)-\omega p(-x)$$and $$vb(x)=p(x)-\omega^2p(-x),$$where $u$ and $v$ are $2$ complex numbers whose product is constant. By looking at the leading coefficients we see that $u$ and $v$ can attain only finitely many values aswell, hence for the sake of our purpose, by infinite PHP we can WLOG that they are fixed. But the above equations imply that $(1-\omega)p(x)=vb(x)-\omega ua(x),$ hence we achieve a contradiction.