Usbrth

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

Find all polynomials $P(x,y)inmathbbR[X,Y]$ for which:

• 
  $P(x,y)$ is homogeneous (so $exists ninmathbbN, forall x,y,tinmathbbR: P(tx,ty)=t^ncdot P(x,y)$).


• $forall a,b,cinmathbbR: P(a+b,c)+P(b+c,a)+P(c+a,b)=0$


• $P(1,0)=1$

My (useful) observations:

1. $P(x,0)=x^n$


2. $P(0,x)=-2x^n$


3. $P(2x,x)=0$


4. $P(x,x)=frac-12cdot2^nx^n$


5. $P(-x,-x)=-P(x,x)= frac12cdot2^nx^n$


6. $P(y,x)=-P(x,y)-(x+y)^n$

When we write
$$P(x,y) = sum_i=0^na_icdot x^iy^n-i$$
These observations imply that:

1. $a_n=1$


2. $a_0=-2$


3. $sum_i=0^na_i=frac-12cdot2^n$


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?

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 $a,binBbbR$ find $cinBbbR$ such that the second identity becomes of the form
$$P(u,-u)+P(v,-v)+P(w,-w)=0.$$

Hint 2:

Deduce that if $degP>1$ 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 $ninBbbN$. 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 $PinBbbR[X,Y]$ with $deg Pleq1$ 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 $a,b,cinBbbR$,
and together with $P(1,0)=1$ this implies $P=X-2Y$.$hspace10ptsquare$

Observation 2: If $PinBbbR[X,Y]$ satisfies the conditions and $deg P>1$ then $X+Y$ divides $P$.

Proof. Suppose $PinBbbR[X,Y]$ satisfies the conditions and $deg P>1$. Plugging in $c=-a-b$ shows that for all $a,binBbbR$ 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 $X+Y$.$hspace10ptsquare$

Proof of full solution. Now we can prove by induction that for all $ninBbbN$ we have

If $PinBbbR[X,Y]$ 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 $ninBbbN$ and suppose that the statement above holds for $n$.

Suppose $PinBbbR[X,Y]$ satisfies the conditions and $deg P=n+2$. Then $P$ is divisible by $X+Y$ by observation 2, which means there exists $QinBbbR[X,Y]$ 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 $a,b,cinBbbR$ we have
  begineqnarray*
  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)),
  endeqnarray*
  which shows that for all $a,b,cinBbbR$ with $a+b+cneq0$ 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 $a,b,cinBbbR$.


• 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
$$Q=(X-2Y)(X+Y)^n
qquadtext and hence qquad
P=(X-2Y)(X+Y)^n+1,$$
which completes the proof by induction.