How to calculate dimesion of homogeneous harmonic polynomial A general homogeneous  polynomial $P:\mathbb{R}^2\to \mathbb{R} $ of degree $n $ is of the form
$$ u\left(x,y\right)=\sum_{k=0}^{n}a_{k}x^{k}y^{n-k} $$
Now, the dimesions of such polynomials would be $n+1 $ with basis $$ x^{0}y^{n},xy^{n-1},...,x^{n}y^{0} $$
I want to find out what is the dimension of such polynomials which are also harmonic, i.e all such polynomials that satisfies $$ \varDelta(\sum_{k=0}^{n}a_{k}x^{k}y^{n-k})=0 $$
I tried to see what a general polynomial need to satisfy and to find condition on $a_k$ that would help me find what is the dimension. It got reallly messy with the algebra and I cant really see where it goes, but here's my attempt:
We want $$ \varDelta u\left(x,y\right)=\sum_{k=0}^{n}\frac{\partial^{2}}{\partial x^{2}}a_{k}x^{k}y^{n-k}+\sum_{k=0}^{n}\frac{\partial^{2}}{\partial y^{2}}a_{k}x^{k}y^{n-k}=0 $$
So
$$ \sum_{k=2}^{n}k\left(k-1\right)a_{k}x^{k-2}y^{n-k}+\sum_{k=0}^{n-2}\left(n-k\right)\left(n-k-1\right)a_{k}x^{k}y^{n-k-2}=0 $$
And now
$$ n\left(n-1\right)a_{n}x^{n-2}+\left(n-1\right)\left(n-2\right)a_{n-1}x^{n-3}y+n\left(n-1\right)a_{0}y^{n-2}+\left(n-1\right)\left(n-2\right)a_{1}xy^{n-3} $$
$$ +\sum_{k=2}^{n-2}k\left(k-1\right)a_{k}x^{k-2}y^{n-k}+\sum_{k=2}^{n-2}\left(n-k\right)\left(n-k-1\right)a_{k}x^{k}y^{n-k-2}=0 $$
So
$$ n\left(n-1\right)\left(a_{n}x^{n-2}+a_{0}y^{n-2}\right)+\left(n-1\right)\left(n-2\right)\left(a_{n-1}x^{n-3}y+a_{1}xy^{n-3}\right) $$
$$ =\sum_{k=2}^{n-2}a_{k}\left(k\left(k-1\right)x^{k-2}y^{n-k}+\left(n-k\right)\left(n-k-1\right)x^{k}y^{n-k-2}\right)=0 $$
Not sure how to proceed. Any ideas would be helpful.
 A: Hint: Instead of trying to think about the whole formula for $\Delta u$ at once, try to think about which terms of $u$ will contribute to each term of $\Delta u$.  Imagine picking the coefficients of $u$ one by one, starting with the coefficient on $x^n$.  What choices are free to make and what choices are forced if you want to make $\Delta u=0$?
More details are hidden below.

 Once you pick the coefficient for $x^n$, that uniquely determines the coefficient on $x^{n-2}y^2$ as well, since you need $\frac{\partial^2}{\partial x^2}$ of the $x^n$ term to cancel out with $\frac{\partial^2}{\partial y^2}$ of the $x^{n-2}y^2$ term.  But then this similarly uniquely determines the coefficient of $x^{n-4}y^4$, which then uniquely determines the coefficient of $x^{n-6}y^6$, and so on.  Similarly, once you've picked the coefficient of $x^{n-1}y$, that determines the coefficients of all the remaining terms with odd powers of $y$.  So (assuming $n>0$) the space of harmonic polynomials is $2$-dimensional, since there are two coefficients you can choose freely to obtain a harmonic polynomial and then the rest are uniquely determined.

