Multivariate polynomial functional equation

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?

• $P(x,y)=(x-2y)(x+y)^{n-1}$ are solutions - but I found this experimentally only
– Sil
Apr 9 '19 at 0:16

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.

• Thanks for the answer! However, there are no solutions for n=0 (P(x,y)=0 doesn’t meet the condition P(1,0)=1). Also, how would you explain your inductory argument? Apr 11 '19 at 9:10
• @JonasDeSchouwer Ah yes, good catch! I'll fix that now, and include the induction argument. Is it ok if I unhide the hidden text while I'm at it? Apr 11 '19 at 9:16
• @JonasDeSchouwer I have added a full proof, with the structure of the induction proof made more explicit. Apr 11 '19 at 9:48
• Thanks! It missed this when I tried to solve the problem: “Because Q is a polynomial, it follows that this holds for all a,b,c \in R. Apr 11 '19 at 10:24

From $$P_n(t x, t y) = t^nP_n(x,y)$$ we conclude that $$P_n(x,y) = \sum_{k=0}^n a_k x^k y^{n-k}$$

now considering

$$P_n(2x,y) + 2P_n(x+y,x) = 0\Rightarrow \sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(x+y\right)^k x^{n-k}$$

or

$$\sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(\sum_{j=0}^k C_j^k x^j y^{k-j}\right) x^{n-k} = \sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(\sum_{j=0}^k C_j^k x^{n-(k-j)} y^{k-j}\right)=0$$

or making $$\nu=k-j$$

$$\sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(\sum_{\nu=k}^0 C_{k-\nu}^k x^{n-\nu} y^{\nu}\right)=0$$

so to guarantee the polynomial identity we conclude

$$2^{n-k}a_{n-k}+2 \sum_{j=0}^{n-k}a_{j+k}C_j^{j+k}= 0$$

with $$a_n = 1$$ due to $$P_n(1,0) = 1$$

This is a almost triangular linear system. For $$n = 3$$ we have

$$\begin{cases} 2^3a_3+2\left(a_0C_0^0+a_1C_1^1+a_2C_2^2+a_3C_3^3\right)=0\\ 2^2a_2+2\left(a_1C_0^1+a_2C_1^2+a_3C_2^3\right)=0\\ 2^1a_1+2\left(a_2C_0^2+a_3C_1^3\right)=0\\ 2^0a_0+2a_3C_0^3 = 0 \end{cases}$$

or

$$\left\{ \begin{array}{rcl} 2 a_0+2 a_1+2 a_2+10 a_3& = &0 \\ 2 a_1+8 a_2+6 a_3&=&0 \\ 2 a_1+2 a_2+6 a_3&=&0 \\ a_0+2 a_3&=&0 \\ \end{array} \right.$$

and solving we have

$$a_0 = -2, a_1 = -3, a_2 = 0, a_3 = 1$$

or

$$P_3(x,y) = (x-2y)(x+y)^2$$

and for $$n$$

$$P_n(x,y) = (x-2y)(x+y)^{n-1}$$

as can be easily verified.

• P(x+y,x) = \sum_{k=0}^{n} a_k (x+y)^{k} x^{n-k} Apr 12 '19 at 9:30
• @JonasDeSchouwer Yes. A typo. Now is corrected. Thanks. Apr 12 '19 at 9:39