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 4 of 4: Confirm every such polynomial works

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P(x)=\beta x^2+\alpha x^4\ (\deg P\le4,\ P(0)=0,\ P\text{ even}) \implies P(a)+P(b)+P(c)=\beta\cdot2s^2+\alpha\cdot2s^4=2P(s)\ \text{whenever } a+b+c=s,\ ab+bc+ca=0

Detailed analysis

By Steps 1 and 3, $P(x)=\beta x^2+\alpha x^4$ for real constants $\alpha,\beta$ . Conversely, let $a,b,c$ with $ab+bc+ca=0$ , and set $x=a-b,y=b-c,z=c-a,s=a+b+c$ , so $x+y+z=0$ and $x^2+y^2+z^2=2s^2$ . Then $xy+yz+zx=-(x^2+y^2+z^2)/2=-s^2$ , so $x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(xy+yz+zx)^2=2s^4$ ; also $x^4+y^4+z^4=2s^4$ , so $P(x)+P(y)+P(z)=\beta\cdot2s^2+\alpha\cdot2s^4$ . Hence $P(x)+P(y)+P(z)=\beta\cdot2s^2+\alpha\cdot2s^4$ equals $2P(s)$ , so this form always works.