Real Analysis 557499.We switch to polar coordinates, where the homogeneity condition becomes the
simpler
u(r, θ )=rng(θ),wheregis a one-variable function of period 2π. Writing the Laplacian = ∂
2
∂x^2 +
∂^2
∂y^2
in polar coordinates, we obtain
=∂^2
∂r^2+
1
r∂
∂r+
1
r^2∂^2
∂θ^2.
For our harmonic function,
0 = u= (rng(θ))=n(n− 1 )rn−^2 g(θ)+nrn−^2 g(θ)+rn−^2 g′′(θ )
=rn−^2 (n^2 g(θ)+g′′(θ )).Therefore,gmust satisfy the differential equationg′′+n^2 g=0. This equation has the
general solutiong(θ)=Acosnθ+Bsinnθ. In order for such a solution to be periodic
of period 2π,nmust be an integer.
(P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics, Springer, 2004)
500.Assume the contrary and writeP(x, y)=(x^2 +y^2 )mR(x, y), whereR(x, y)is not
divisible byx^2 +y^2. The harmonicity condition can be written explicitly as
4 m^2 (x^2 +y^2 )m−^1 R+ 2 m(x^2 +y^2 )m−^1(
x∂R
∂x+y∂R
∂y)
+(x^2 +y^2 )m(
∂^2 R
∂x^2+
∂^2 R
∂y^2)
= 0.
IfR(x, y)weren-homogeneous for somen, then Euler’s formula would allow us to
simplify this to
( 4 m^2 + 2 mn)(x^2 +y^2 )m−^1 R+(x^2 +y^2 )m(
∂^2 R
∂x^2+
∂^2 R
∂y^2)
= 0.
If this were true, it would imply thatR(x, y)is divisible byx^2 +y^2 , a contradiction.
But the polynomialx^2 +y^2 is 2-homogeneous andR(x, y)can be written as a sum
ofn-homogeneous polynomials,n= 0 , 1 , 2 ,....Since the Laplacian∂x∂ 2 +∂y∂ 2 maps
ann-homogeneous polynomial to an(n− 2 )-homogeneous polynomial, the nonzero
homogeneous parts ofR(x, y)can be treated separately to reach the above-mentioned
contradiction. HenceP(x, y)is identically equal to zero.
Remark.The solution generalizes in a straightforward manner to the case ofnvariables,
which was the subject of a Putnam problem in 2005. But as I.M. Vinogradov said, “it is
the first nontrivial example that counts.’’