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 3 of 4: Bound the degree using leading coefficients

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\deg P=n\ge6,\ \text{leading coeff }c_n\ne0 \implies c_n(3^n+5^n+8^n)=2c_n7^n \implies (8/7)^n\ge2 \text{ for } n\ge6 \Rightarrow \text{contradiction}

Detailed analysis

Suppose $P$ is not identically zero, of degree $n$ with leading coefficient $c_n\ne0$ . Comparing the coefficient of $t^n$ on both sides of $P(3t)+P(5t)+P(8t)=2P(7t)$ gives $3^n+5^n+8^n=2\cdot7^n$ . Since $P$ is even and $P(0)=0$ , $n$ is even. For $n\ge6$ , $(8/7)^n$ already exceeds $2$ , hence $3^n+5^n+8^n=2\cdot7^n$ forces $n\ge6$ ; checking directly, $n=2$ and $n=4$ both satisfy $3^n+5^n+8^n=2\cdot7^n$ , so $n$ can only be $2$ or $4$ .