MathLabs

Problem 6

Find all polynomials PP in two variables such that, for a positive integer nn, P(tx,ty)=tnP(x,y)P(tx,ty)=t^nP(x,y) for all real t,x,yt,x,y; for all real a,b,ca,b,c, P(b+c,a)+P(c+a,b)+P(a+b,c)=0P(b+c,a)+P(c+a,b)+P(a+b,c)=0; and P(1,0)=1P(1,0)=1.
Step 2 of 6: Find a linear factor
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3P(2u,u)=0  ⟹  P(2u,u)=0  ⟹  (x−2y)∣P(x,y)3P(2u,u)=0\implies P(2u,u)=0\implies (x-2y)\mid P(x,y)
Detailed analysis

Set a=b=c=ua=b=c=u in the cyclic identity. Then 3P(2u,u)=03P(2u,u)=0 for every uu. Thus PP vanishes on x=2yx=2y, so x−2yx-2y divides PP.