Problem 2

Find all polynomials

P

with real coefficients such that for all real numbers

a,b,c

with

ab+bc+ca=0

, we have

P(a-b)+P(b-c)+P(c-a)=2P(a+b+c).

Step 1 of 4:

P

is even with

P(0)=0

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a=b=c=0 \implies P(0)=0;\quad b=c=0 \implies P(a)+P(0)+P(-a)=2P(a) \implies P(-a)=P(a)

Detailed analysis

Taking $a=b=c=0$ (which satisfies $ab+bc+ca=0$ ) gives $3P(0)=2P(0)$ , so $P(0)=0$ . Taking $b=c=0$ (again satisfying $ab+bc+ca=0$ automatically) turns the equation into $P(a)+P(0)+P(-a)=2P(a)$ , and since $P(0)=0$ , this simplifies to $P(-a)=P(a)$ for every real $a$ , so $P$ is an even function.