Multivariate polynomial functional equation

functional-equationshomogeneous equationmultivariate-polynomial

I’m having some difficulties solving the following functional equation:

Find all polynomials $P(x,y)\in\mathbb{R}[X,Y]$ for which:

  • $P(x,y)$ is homogeneous (so $\exists n\in\mathbb{N}, \forall x,y,t\in\mathbb{R}: P(tx,ty)=t^n\cdot P(x,y)$).
  • $\forall a,b,c\in\mathbb{R}: P(a+b,c)+P(b+c,a)+P(c+a,b)=0$
  • $P(1,0)=1$

My (useful) observations:

  1. $P(x,0)=x^n$
  2. $P(0,x)=-2x^n$
  3. $P(2x,x)=0$
  4. $P(x,x)=\frac{-1}{2}\cdot2^nx^n$
  5. $P(-x,-x)=-P(x,x)= \frac{1}{2}\cdot2^nx^n$
  6. $P(y,x)=-P(x,y)-(x+y)^n$

When we write
$$P(x,y) = \sum_{i=0}^na_i\cdot x^iy^{n-i}$$
These observations imply that:

  1. $a_n=1$
  2. $a_0=-2$
  3. $\sum_{i=0}^na_i=\frac{-1}{2}\cdot2^n$
  4. $\sum_{i=0}^n2^ia_i=0$

Also, the fifth observation implies that $n$ is odd.

I’ve noticed that $P(x,y)=x-2y$ satisfies the conditions, but I don’t know how to prove it’s the only solution.

Can someone please give me a hint how to proceed?

Best Answer

As you ask for a hint, here are two hints to help you along. Below is a sketch of a full proof. Let me know when I can 'unhide' all the hidden text to make the answer more legible for future readers.

Hint 1:

For all $a,b\in\Bbb{R}$ find $c\in\Bbb{R}$ such that the second identity becomes of the form $$P(u,-u)+P(v,-v)+P(w,-w)=0.$$

Hint 2:

Deduce that if $\deg{P}>1$ then $P$ is divisible by $X+Y$.

Full solution: The polynomials that satisfy the conditions are precisely the polynomials $$(X-2Y)(X+Y)^n,$$ with $n\in\Bbb{N}$. It is not hard to verify that these polynomials satisfy the conditions. Showing that there are no other solutions is more work. Below is a proof is by induction on the degree.

Observation 1: The unique solution $P\in\Bbb{R}[X,Y]$ with $\deg P\leq1$ is $P=X-2Y$.

Proof. There are no constant solutions, and for $n=1$ setting $P=uX+vY$ shows that $$(2u+v)(a+b+c)=0,$$ holds for all $a,b,c\in\Bbb{R}$, and together with $P(1,0)=1$ this implies $P=X-2Y$.$\hspace{10pt}\square$

Observation 2: If $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P>1$ then $X+Y$ divides $P$.

Proof. Suppose $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P>1$. Plugging in $c=-a-b$ shows that for all $a,b\in\Bbb{R}$ we have $$0=P(a+b,-a-b)+P(-a,a)+P(-b,b)=((a+b)^n+(-a)^n+(-b)^n)P(1,-1),$$ which implies that $P(1,-1)=0$ because $n>1$, and hence that $P(X,-X)=0$. This means $P$ is divisible $X+Y$.$\hspace{10pt}\square$

Proof of full solution. Now we can prove by induction that for all $n\in\Bbb{N}$ we have

If $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P=n+1$ then $P=(X-2Y)(X+Y)^n$.

The base case $n=0$ is covered by observation 1. So let $n\in\Bbb{N}$ and suppose that the statement above holds for $n$.

Suppose $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P=n+2$. Then $P$ is divisible by $X+Y$ by observation 2, which means there exists $Q\in\Bbb{R}[X,Y]$ such that $P=(X+Y)Q$. Then clearly $\deg Q=n+1$, and we verify that $Q$ also satisfies the conditions:

  • Because $P$ and $X+Y$ are homogeneous, also $Q$ is homogeneous.
  • For all $a,b,c\in\Bbb{R}$ we have \begin{eqnarray*} 0&=&P(a+b,c)+P(b+c,a)+P(c+a,b)\\ &=&(a+b+c)(Q(a+b,c)+Q(b+c,a)+Q(c+a,b)), \end{eqnarray*} which shows that for all $a,b,c\in\Bbb{R}$ with $a+b+c\neq0$ we have $$Q(a+b,c)+Q(b+c,a)+Q(c+a,b)=0.$$ Because $Q$ is a polynomial, it follows that this holds for all $a,b,c\in\Bbb{R}$.
  • Clearly $P(1,0)=1$ implies $Q(1,0)=1$.

This shows that $Q$ satisfies the conditions and $\deg Q=n+1$, so by induction hypothesis $$Q=(X-2Y)(X+Y)^n \qquad\text{ and hence }\qquad P=(X-2Y)(X+Y)^{n+1},$$ which completes the proof by induction.

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