Find all polynomials $p(x)$ with real coeffcients such that \[p(a + b - 2c) + p(b + c - 2a) + p(c + a - 2b) = 3p(a - b) + 3p(b - c) + 3p(c - a)\] for all $a, b, c\in\mathbb{R}$. (2nd Benelux Mathematical Olympiad 2010, Problem 2)
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Tags: algebra, polynomial, functional equation, algebra proposed
02.05.2010 20:31
Putting $x:=a-b,y:=b-c$, this becomes $p(x-y)+p(-2x-y)+p(x+2y)=3p(x)+3p(y)+3p(-x-y)$ for all real $x,y\qquad(*)$ For $x=y=0$ we get $3p(0)=9p(0)\iff p(0)=0$ For $y=0$ we get $2p(x)+p(-2x)=3p(x)+3p(-x)\iff p(-2x)=p(x)+3p(-x)\qquad(**)$ If $p(x)=a_nx^n+\dots+a_2x^2+a_1x$, then from $(**)$ $a_k(-2)^k=a_k+3a_k(-1)^k,k=\overline{1,n}$ $a_k\left[(-2)^k-1-3(-1)^k\right]=0\iff \begin{cases}a_k(2^k-4)=0 & 2\mid k\\ a_k(2^k-2)=0 & 2\nmid k\end{cases}$ For $k\geqslant 3$, the equations yield $a_k=0$. Therefore $p(x)=\lambda x^2+\mu x$, where $\lambda, \mu$ are real constants. Plugging that back into $(*)$, we find no constraints on $\lambda,\mu$. Therefore $p(x)=\lambda x^2+\mu x, \lambda,\mu\in\mathbb{R}$
25.09.2010 07:39
just put a=b thensee the equation reduces to $\ p(2b-2c)+p(c-b)+p(c-b)=3p(0)+3p(b-c)+3p(c-b) $ as b,c can be arbitrary so we caqn put x=b-c where x can be arbitrary. then it gives a functional equation of following type $ \ p(2x)=3a_{0}+3p(x)+p(-x) $ where $ \ p(x)=a_{n}x^{n}+......a_{1}x+a_{0} $ so for n>=1 $ \ 2^{n}a_{n}=3a_{n}+(-1)^{n}a_{n} $ so see for k>=3 $ \ a_{k}=0$ also see $ \ a_{0}=0 $ now just put back $ \ p(x)=\alpha x^{2}+\beta x $ for arbitrary $ \alpha, \beta $ in the first given equation to see it gives an identity. so done.
06.02.2025 14:31
The polynomials are all \(p(x) = dx^2 + ex\) for all \((d,e) \in \mathbb{R}^2\). It's easy to see that \[\sum_{cyc} d{(a+b-2c)}^2 = \sum_{cyc} 3d{(a-b)}^2 = 6d(a^2+b^2+c^2-ab-bc-ca)\]We also have \[\sum_{cyc} e(a+b-2c) = \sum_{cyc} 3e(a-b) = 0\]Thus all polynomials of this form work. Next we shall prove that all polynomials of degree \(3\) or higher fail. We first claim $p(0) = 0$ (and thus our polynomial cannot have a constant term), indeed this is obvious upon substituting $(a,b,c) = (0,0,0)$, from which we obtain \[3f(0) = 9f(0) \Longrightarrow f(0) = 0\]Notice that by substituting \((a,b,c) = (a,0,0)\) we can obtain the equation \[2p(-a) + p(2a) = 3p(a) + 3p(-a)\]Let \(a_k\) be the leading coefficient of this polynomial, since we must have the leading coefficient be equal in both the L.H.S and R.H.S (letting \(n\) denote the degree of the polynomial) we obtain the equation \[2a_k{(-1)}^n + a_k \cdot 2^n = 3a_k + 3a_k{(-1)}^n\]It is easy to see that the equation does not hold for \(n=3\) and that for \(n \geq 4\), we must have the L.H.S be at least \(14\) whereas the R.H.S is at most \(6\), thus \(n \geq 4\) doesn't hold either. Hence we are done.