1975 IMO Problems/Problem 6

Problem

Find all polynomials $P$, in two variables, with the following properties:

(i) for a positive integer $n$ and all real $t, x, y$ $$P(tx, ty) = t^nP(x, y)$$ (that is, $P$ is homogeneous of degree $n$),

(ii) for all real $a, b, c$, $$P(b + c, a) + P(c + a, b) + P(a + b, c) = 0,$$

(iii) $$P(1, 0) = 1.$$

Solution 1

(i) If $n = 0$ : Clearly no solution (ii) If $n = 1$ : $P(x, y) = ax+by \implies$ the identification yields directly $P(x,y) = x-2y$ (iii) If $n > 1$, $a = 1, \ b = 1, \ c =-2 \implies P(2,-2)+P(-1, 1)+P(-1, 1) = 0$ $\implies ((-2)^{n}+2) P(-1, 1) = 0 \implies P(-1, 1) = 0 \implies P(-y, y) = 0 \implies$ $P(x,y)$ is divisible by $(x+y)$ It is then easy to see that $\frac{P(x, y)}{(x+y)}$, of degree $n-1$ verifies all the equations.

The only solutions are thus $P(x, y) = (x-2y)(x+y)^{n-1}$

The above solution was posted and copyrighted by mathmanman. The original thread for this problem can be found here: [1]

Solution 2

$F(a,a,a) \implies P(2a,a)=0 \implies (x-2y)$ is a factor of $P(x,y)$.

We may write $P(x,y)=(x-2y)Q(x,y)$

$F(a,b,b) \implies 2P(a+b,b)+P(2b,a)=0 \implies 2(a+b-2b)Q(a+b,b)+2(b-a)Q(2b,a) \implies (a-b)(Q(a+b,b)-Q(2b,a))=0$ Thus $Q(a+b,b)=Q(2b,a) \forall a \neq b$

We may rewrite it as $Q(x,y)=Q(2y,x-y)=Q(2x-2y,3y-x=\cdots$ $Q(x+d,y-d)-Q(x,y)$ is a polynomial in $d$ of degree $n-1$ for any two fixed $x,y$,which has infinitely many zeroes,i.e,$0,2y-x,x-2y,6y-3x,\cdots$.Thus $Q(x+d,y-d)=Q(x,y)$ holds for all $d$.In particular it holds for $d=y$,i.e, $Q(x+y,0)=Q(x,y)$.Now consider the polynomial $R(x,y)=Q(x+y,0)-Q(x,y)$.Suppose that its not the zero polynomial.Then its degree $d$ is defined.With $t=\frac{x}{y}$ it can be wriiten as $y^dS(t)=y^d(A(t)-B(t))$.But $S(t)$ has infinitely many zeroes and this forces $A(t)=B(t)$,forcing $R(x,y)$ to be a zero polynomial.Contradiction!.Thus $Q(x+y,0)$ and $Q(x,y)$ are identical polynomials.This forces $Q(x,y)=c(x+y)^{n-1}$.With $Q(1,0)=1$ we get $c=1$.Thus $$P(x,y)=(x-2y)(x+y)^{n-1}$$

The above solution was posted and copyrighted by JackXD. The original thread for this problem can be found here: [2]