Determine all polynomials $P(x,y)$ with real coefficients such that \[P(x+y,x-y)=2P(x,y) \qquad \forall x,y\in\mathbb{R}.\]
Problem
Source:
Tags: algebra, polynomial, algebra proposed
15.08.2012 23:51
12.02.2014 09:13
peregrinefalcon88 wrote: Now examining (1) tells us that the degree of any term containing x's and y's is at most 2 since we know that the degree of the x-only-terms is 2. We now know that Why? In fact for any polynomial $P(x,y)=x^2+x^2y^2+y^2$ we have also have leading powers to be $2$ for $P(x,0)=x^2$ and for $P(0,y)=y^2$. Someone please try to solve this problem or post official solutions.
12.02.2014 18:18
Uh... Did you look at equation 1? Your example doesn't work, because $P(x, x)$ is then $x^4+2x^2$, of degree $4$.
12.02.2014 22:41
The step hal9v4ik highlighted is definitely suspect, though: for example, the polynomial $P(x, y) = x^3y - xy^3$ satisfies all the numbered equations (1) through (6) that peregrinefalcon88 wrote down, but has terms of degree larger than 2. So certainly the derivation of (7) (and so the solution) is wrong. Edit: okay, and here's a much more direct and less wrong way to get to the same place: we have \begin{align*} P(2x, 2y) & = P((x + y) + (x - y), (x + y) - (x - y)) \\ & = 2P(x + y, x - y) \\ & = 4P(x, y), \end{align*} so $P$ is homogeneous of degree 2. The last steps are routine.
13.02.2014 04:45
Okay,I also got this but never thought homogenity for polynomial with two variables.
04.09.2014 07:52
I'm sorry JBL ,but your example is wrong,because if $ P(x,y)=x^3y-y^3x=xy(x-y)(x+y) $ , then $ LHS=4xy(x-y)(x+y) $ ,but $ RHS=2xy(x-y)(x+y) $.
04.09.2014 17:52
Sardor, that was his point. It of course doesn't satisfy the equation, because as he writes later in his same post, that any solution must be homogeneous of degree 2, which it is not. His point is that peregrinefalcon88's proposed solution method was wrong because it never explicitly ruled out $x^3y-xy^3$ as a solution.
09.01.2025 21:10
Hello,can someone provide correct solution?