I want to determine $I(V)$ (the ideal of $V$) for $V(x+y)=\{(x,y)\in\mathbb{R}^2: x=-y\}$. It is defined as $I(V)=\{f\in\mathbb{R}[x,y]:f(a,b)=0 \forall (a,b)\in V\}$. So e.g. we have $f_1,f_2,f_3\in V$ for $f_1(x,y)=x+y, f_2(x,y)=x^3+y^3, f_3(x,y)=x^3+y$. But I do not really get the pattern... Any help is greatly appreciated!
Edit: $f_3$ is not in $V$!
HINT: the polynomials $x^n + y^n$ you already found, all have a nice decomposition:
$$(x^n + y^n) = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \ldots - xy^{n-2} + y^{n-1})$$
Similiarly we find that $x^2 - y^2 \in I(V)$ thanks to the decomposition
$$x^2 - y^2 = (x + y)(x-y)$$
Second hint: knowing that both $x^3 + y^3$ and $x^5 + y^5$ are in $I(V)$ you get for free that $x^5 + 2x^3 + 2y^3 + y^5$ is as well.
All in all $I(V)$ is a lot larger than you think (in the comment).