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 2 of 4: Plug in a scaling family

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(a,b,c)=(6t,3t,-2t):\ ab+bc+ca=18t^2-6t^2-12t^2=0;\ a-b=3t,\ b-c=5t,\ c-a=-8t,\ a+b+c=7t

Detailed analysis

For any real $t$ , the triple $(a,b,c)=(6t,3t,-2t)$ satisfies $ab+bc+ca=0$ since $18t^2-6t^2-12t^2=0$ . Substituting into the functional equation and using that $P$ is even (so $P(-8t)=P(8t)$ ) turns it into $P(3t)+P(5t)+P(8t)=2P(7t)$ , an identity in $t$ for every $P$ satisfying the equation.