amparvardi wrote: Find all polynomials $f(x)$ with real coefficients for which \[f(x)f(2x^2) = f(2x^3 + x).\] Here is a rather ugly solution. I hope somebody will find a better one. The only constant solutions are $f(x)=0$ and $f(x)=1$ $\forall x$. Let us from now look for non constant solutions. Let $P(x)$ be the assertion $f(x)f(2x^2)=f(2x^3+x)$ $P(0)$ $\implies$ $f(0)^2=f(0)$ and so $f(0)=0$ or $f(0)=1$ If $f(0)=0$ : let $f(x)=x^pg(x)$ with $g(0)\ne 0$. The assertio $p(x)$ becomes $2{}^px{}^{3p}g(x)g(2x^2)=x^p(2x^2+1)^pg(2x^3+x)$ and so (remember these are polynomials) : $2{}^px{}^{2p}g(x)g(2x^2)=(2x^2+1)^pg(2x^3+x)$ which is impossible (just set $x=0$ in it) So $f(0)=1$ It's easy to see that $f(a)=0$ implies $f(2a^3+a)=0$ and so infinitely many roots if $a\ne 0$. So $f(x)$ has no real roots, and so $f(x)>0$ $\forall x$ $P(i)$ $\implies$ $f(i)f(-2)=f(-i)$ $P(-i)$ $\implies$ $f(-i)f(-2)=f(i)$ And so $f(i)f(-2)^2=f(i)$ If $f(-2)^2\ne 1$, this implies $f(i)=f(-i)=0$ and let then $f(x)=(x^2+1)^pg(x)$ with $g(i)\ne 0$ Assertion $P(x)$ becomes $(x^2+1)^p(4x^4+1)^pg(x)g(2x^2)=((2x^3+x)^2+1)^pg(2x^3+x)$ $=(x^2+1)^p(4x^4+1)^pg(2x^3+x)$ and so $g(x)g(2x^2)=g(2x^3+x)$ So any non constant solution may be written as $(x^2+1)^pf(x)$ where $f(x)$ is a solution such that $f(i)\ne 0$ So we got another family of solutions : $(x^2+1)^n$ Let us from now look for non constant solutions $f(x)$ such that $f(i)\ne 0$ $f(i)f(-2)^2=f(i)$ implies then $f(-2)=1$ (since we know that $f(x)>0$ $\forall x$ Let then $x_m$ such that $f(x_m)$ is a global minimum for $f(x)$. We get $f(x_m)\le f(0)=1$ If $f(x_m)=1$, choose $x_m=-2$ and let then $u$ such that $2u^3+u=x_m=-2$. we get $-2<u<0$ and $P(u)$ $\implies$ $f(u)f(2u^2)=f(x_m)=1$ and so $f(u)=f(2u^2)=1$. Let then $u_1$ such that $2u_1^3+u_1=u$. We get $u<u_1<0$ and, in the same way, $f(u_1)=1$ And so we can buils infinitely many $u_i$ such that $f(u_i)=1$, which is impossible. So $f(x_m)<1$ and $x_m\ne 0$ Let then $u\ne 0$ such that $2u^3+u=x_m$ : $P(u)$ $\implies$ $f(u)f(2u^2)=f(x_m)\le\min(f(u),f(2u^2)$ and so $f(u)\le 1$ and $f(2u^2)\le 1$ $P(x_m)$ $\implies$ $f(x_m)f(2x_m^2)=f(2x_m^3+x_m)\ge f(x_m)$ and so $f(2x_m^2)\ge 1$ So $f(2u^2)\le 1$ and $f(2{}_x{}_m^2)\ge 1$ and so $\exists v_1>0$ such that $f(v_1)=1$ Let then $v1> w_1>0$ such that $2w_1^2+w_1=v_1$ $P(w_1)$ $\implies$ $f(w_1)f(2w_1^2)=f(v_1)=1$ $v_1=2w_1^3+w_1>2w_1^2$ so that both $w_1$ and $2w_1^2$ are in $(0,v_1)$ And since $f(w_1)f(2w_1^2)=1$, one of the two values is $\ge 1$ and the other is $\le 1$ and so $\exists v_2\in(0,v_1)$ such that $f(v_2)=1$ And so we can build an infinite decreasing sequence of positive real numbers such that $f(v_i)=1$, which is impossible. And so there are no non constant solutins such that $f(i)\ne 0$ So the only solutions are : $f(x)=0$ $f(x)=1$ $f(x)=(x^2+1)^n$ with any $n\in\mathbb N$

This problem is ISL 1979 Problem 3. And Mr. Patrick your solution is exactly the same as official solution, congrats

The only solutions are $f(x) = 0$ and $f(x) = (x^2 + 1)^n$ for some nonnegative integer $n$. Setting $x=0$ yields $f(0) = 0$ or $f(0) = 1$. We claim that if $f(0) = 0$, then $f$ is the zero polynomial. Suppose for the sake of contradiction that $f$ is not the zero polynomial. Then we may let $f(x) = x^n g(x)$, where $g(0) \neq 0$. It follows that $(2x^3)^n g(x)g(2x^2) = (2x^3 + x)^n g(2x^3 + x)$, so $(2x^2)^n g(x)g(2x^2) = (2x^2 + 1)^n g(2x^3 + x)$. Setting $x = 0$ yields $0 = g(0)$, which is a contradiction. We now consider the case in which $f(0) = 1$. We observe that for any solution $h(x)$ to the functional equation, $(x^2+1) h(x)$ is also a solution. Hence, it is sufficient to solve this equation in the case in which $x^2+1$ does not divide $f$. We claim that if $x^2+1$ does not divide $f$, then $f$ must be constant. Suppose for the sake of contradiction that it is nonconstant. Let its roots be $z_1, z_2, \ldots, z_m$, and suppose without loss of generality that $z_1$ is any of the roots with greatest magnitude. Because $f(0) = 1$, by Vieta's formulas $|z_1| |z_2| \cdots |z_m| = 1$. Hence, $|z_1| \geq 1$. Setting $x = z_1$ in the equation yields $0 = f(z_1) f(2z_1^2) = f(2z_1^3 + z_1)$, so $2z_1^3 + z_1$ is a root as well. Because $z_1$'s norm is maximal, $|z_1| \geq |2z_1^3 + z_1|$, so $|2z_1^2 + 1| \leq 1$. However, $2 \geq |-2z_1^2 - 1| + |1| \geq |-2z_1^2 - 1 + 1| = 2|z_1|^2 \geq 2$. From the equality case of the triangle inequality, $z_1^2$ must be a negative real. Since $|z_1| = 1$, $z_1^2 = \pm i$. Hence, either $i$ or $-i$ must be a root. But if $i$ is a root, then $0 = f(i)f(-2) = f(2i^3 + i) = f(-i)$, and if $-i$ is a root, then $0 = f(-i)f(-2) = f(i)$, so if either $i$ or $-i$ is a root, then both $i$ and $-i$ are roots. This implies that $x^2+1$ is a factor of $f$, which is a contradiction. It follows that $f$ is constant. It is easy to conclude from here that the only solutions are $f(x) = 0$ and $f(x) = (x^2 + 1)^n$ for any nonnegative integer $n$.

My solution from WOOT: $x=0$ gives $f(0)\in\{0,1\}$ Case 1: $f(0)=0$ Here $\boxed{f(x)=0}$ works, otherwise let $f(x)=x^ng(x)$ with $g(0)\ne0$. We have (after dividing through by $x^n$): $$2^nx^{2n}g(x)g(2x^2)=(2x^2+1)^ng(x)$$Setting $x=0$, we have $g(0)=0$, contradiction. Case 2: $f(0)=1$ Notice that if $f(x)$ fits, then $(x^2+1)f(x)$ does also. With this, let $f(x)=(x^2+1)^ng(x)$ with $x^2+1\nmid g(x)$. We have $g(x)g(2x^2)=g(2x^3+x)$ and $g(0)=1$. Let $r$ be a root of $f$. Notice $r\ne0$. Let the sequence $a_n$ be defined by $a_1=r$ and $a_{n+1}=2a_n^3+a_n$. Assume first that $|r|>1$. We claim that $a_{n+1}>a_n$ for all $i$. Assume for the sake of induction that $|a_n|>|a_{n-1}|>1$ for some $n$ (the base case is easy with the same method). Since $|2a_n^3+a_n|+|-a_n|\ge|2a_n^3|$ by the triangle inequality we have: $$|a_{n+1}|=|2a_n^3+a_n|\ge|2a_n^3|-|a_n|=|a_n|\left(2|a_n|^2-1\right)>|a_n|.$$Thus all of the $a_n$ are distinct, so $g$ has infinitely many roots, leading to $g(x)=0$, contradiction. If $|r|<1$, we obtain a similar contradiction. Thus all roots $r$ satisfy $|r|=1$. But then $|2r^3+r|\ge|2r^3|-|r|=2|r|^3-|r|=1$, so equality holds and $2r^3=kr$ for some $k\in\mathbb R$, or $2r^2=k$. Thus, $\deg g\le2$. There are no suitable solutions for $g$ with a root. The only alternative is that $g$ is constant with $g\not\equiv0$. This gives $g(x)=1$ and $\boxed{f(x)=(x^2+1)^n},n\in\mathbb N_0$.

The only such functions are $\boxed{f(x) = 0}$ and $\boxed{f(x) = (x^2+1)^m}$ for some integer $m$, which both work. Claim. $f(x)=0$ or $f(0)=1$. Plug in 0 to get $f(0) = 0$ or $f(0) =1$. Suppose $f(0)=0$ -- then $x$ is a factor of $f(x)$, so set $f(x) = x^kg(x)$ for some $g(0) \neq 0$. The assertion implies $$x^kg(x) \cdot 2^{2k}x^{2k}g(2x^2) = (2x^3+x)^kg(2x^3+x) \iff 2^k(x^{2k}) g(x)g(x^2) = (2x^2+1)^kg(2x^3+x).$$However, setting $x=0$ in this equation gives us $g(0)=0$, a contradiction. THus the only possibility if $f(0)=0$ is to have $f(x) = 0$. $\blacksquare$ Claim. All roots $r$ of $f(x)$ satisfy $|r| = 1$. Suppose $|r| \neq 1$. By the equation, if $r$ is a root, then $2r^3+r$ is also a root. But $$|r| > 1 \implies |2r^3+r| > 2|r|^3-|r| > 1,$$leading to infinitely many solutions, so $|r| \leq 1$. However, the constant term $f(0)=1$ by the previous claim, so we must have $|r| = 1$ for all $r$, as required. $\blacksquare$ Now we can explicitly compute the roots, as we will see: Claim. Only $\pm i$ can be roots of $f(x)$. Observe that to prevent infinite solutions, $$|2r^3+r| = |r| \iff |2r^2+1| = 1,$$which expands to $$4(r\overline r)^2+2(r^2+\overline r^2) + 1 = 1.$$However, $r \overline r = 1$, so $$2(r^2+\overline r^2) = -4, r^2+\overline r^2 = -2.$$This means that $r+\overline r= 0$, so $r$ is pure imaginary. It follows $r = \pm i$, as required. $\blacksquare$ Thus, $x^2+1$ is the only possible factor of $f$, and we are done.

If \( P \) is a constant, then clearly \( P \equiv 0 \) or \( P \equiv 1 \). Now let \( P(x) = \sum_{k=0}^n a_k x^k \), where \( a_n \neq 0 \) and \( n \geq 1 \). Then the degree of \( P(2x^2) \) is \( 2n \), and the leading coefficient is \( 2^n a_n \), while the degree of \( P(2x^3 + x) \) is \( 3n \), and the leading coefficient is also \( 2^n a_n \). Thus, comparing the leading coefficients on both sides of the given equation, we obtain \( 2^n a_n^2 = 2^n a_n \), i.e., (since \( a_n \neq 0 \)) it follows that \( a_n = 1 \). Now we calculate \( a_0 \). Let \( P(x) = x^k Q(x) \), where \( Q(0) \neq 0 \) and \( k \geq 0 \). Then for \( x \neq 0 \), \[ x^k Q(x)(2x^2)^k Q(2x^2) = (2x^3 + x)^k Q(2x^3 + x) \Leftrightarrow 2^k x^{2k} Q(x) Q(2x^2) = (2x^2 + 1)^k Q(2x^3 + x). \]Since the last equality holds for all \( x \neq 0 \), it also holds for \( x = 0 \) (the difference between the two sides is a polynomial with infinitely many roots and hence identically zero). Thus, for \( k \geq 1 \) and \( x = 0 \), we get \( Q(0) = 0 \), a contradiction. Therefore, \( k = 0 \), and in particular, \( P(0) = Q(0) \neq 0 \). Now, setting \( x = 0 \) in the given equation gives \( P(0)^2 = P(0) \), from which \( a_0 = P(0) = 1 \). Determining \( a_0 \) and \( a_n \) helps us with the following — we now have information about the product of the complex roots of \( P \): from Vieta’s formulas, it is \( (-1)^n \frac{a_0}{a_n} \). In particular, the product of the magnitudes of the roots of \( P \) is 1. Now let \( \alpha \) be a root of \( P \) with the maximum modulus. Since the product of the moduli of the roots is $1$ by Vieta's formulae, it follows that \( |\alpha| \geq 1 \). Substituting \( x = \alpha \) in the given equation gives \( P(2\alpha^3 + \alpha) = 0 \), i.e., \( 2\alpha^3 + \alpha \) is also a root of \( P \). From the maximality of \( |\alpha| \), we have \( |2\alpha^3 + \alpha| \leq |\alpha| \), or equivalently, \( |2\alpha^2 + 1| \leq 1 \). Using the triangle inequality in the form \( |a + b| \geq |a| - |b| \), it follows that \[ 1 \geq |2\alpha^2 + 1| \geq |2\alpha^2| - 1 = 2|\alpha|^2 - 1, \]and hence \( |\alpha| \leq 1 \). Therefore, \( |\alpha| = 1 \). Since equality holds everywhere above, it follows that \( |2\alpha^2 + 1| = 1 \). Writing \( \alpha = a + bi \), we have \( 1 = |\alpha|^2 = a^2 + b^2 \) and \[ 1 = |2\alpha^2 + 1|^2 = (2(a^2 - b^2) + 1)^2 + 16a^2b^2. \]From this, \( b^2 = 1 - a^2 \), and substituting gives \[ 1 = (4a^2 - 1)^2 + 16a^2(1 - a^2) = 8a^2 + 1, \]which implies \( a = 0 \) and \( b = \pm 1 \). In particular, at least one of the numbers \( i \) and \( -i \) is a root of \( P \). However, from the well-known fact that if \( \beta \) is a root of a polynomial with real coefficients, then its conjugate \( \overline{\beta} \) is also a root, it follows that both \( i \) and \( -i \) are roots of \( P \). Thus, we have \( P(x) = (x^2 + 1)R(x) \), where \( R(x) \) is a polynomial with real coefficients. Substituting this into the original equation gives \[ (x^2 + 1)R(x)(4x^4 + 1)R(2x^2) = ((2x^3 + x)^2 + 1)R(2x^3 + x) \Leftrightarrow R(x)R(2x^2) = R(2x^3 + x) \]for all real \( x \). This equation is the same as the original one (except with \( R \) in place of \( P \)), and thus by induction it follows that \( P(x) = (x^2 + 1)^n \) for some non-negative integer \( n \). Conversely, a similar calculation shows that the polynomials \( 0 \) and \( (x^2 + 1)^n \) for \( n \geq 0 \) satisfy the given condition.