real affine varieties are hypersurfaces 
In $\mathbb{R}^n$, let X be a Zariski-closed set. then $X=\mathbb{V}(f)$ for some polynomial $f$.

Elementary formulation: let $X \subset \mathbb{R}^n$ be the set of common zeroes of some polynomials $f_i$ in $\mathbb{R}[x_1,...,x_n]$, prove there is a polynomial $f\in \mathbb{R}[x_1,...,x_n]$ such that $X$ is the set of zeroes of $f$.
This is of course false replacing $\mathbb{R}$ with $\mathbb{C}$. Passing to an ideal formulation seems problematic since we don't have the nullstellensatz, and hence $I(X)=I(\mathbb{V}(f))$ doesn't simplify.
The case $n=1$ is easy since the closed proper subsets of $\mathbb{R}$ are finite and we can take explicitly a polynomial that has as roots exactly $X$.
We can assume $X$ is irreducible since then we use it for each irreducible component and get $X=\cup _{i=1}^r X_i=\cup \mathbb{V}(f_i)=\mathbb{V}(\prod f_i)$.
 A: Real numbers are bizarre. Since any such $X$ is defined by finitely many $f_i$'s, just take $f=∑f_i^2$.
A: Actually, this holds for all non algebraically closed fields:

*

*Affine variety over a field which is not algebraically closed can be written as the zero set of a single polynomial.

Edit: Here is Eric Wolfsey's argument in the link above (I am just copying it here for convenience, but I want to make it clear that I learned this from his post, appearing in block quotes below).
First, note that in order to show that an algebraic subset $\mathcal{V}(f_1,\,\ldots\,,f_m)\subset k^n$ can be written under the form $\mathcal{V}(g)$ for a single $g\in k[X_1,\,\ldots\,,X_n]$, it suffices to find a function $\phi\in k[X_1,\,\ldots\,,X_m]$ such that $\mathcal{V}(\phi)=\{0\}$, and then set $g=\phi(f_1,\,\ldots\,,f_m)$.

And it actually suffices to find such a $\phi$ in the case $m=2$. For instance, if you have such a $\phi$ for $m=2$, then $\phi(X_1,\phi(X_2,X_3))$ works for $m=3$, and $\phi(X_1,\phi(X_2,\phi(X_3,X_4)))$ works for $m=4$, and so on.

If $k=\mathbb{R}$, you can just take $\phi(X,Y) = X^2 + Y^2$. And if $k$ is an arbitrary non algebraically closed field:

what you can do is take a nonconstant polynomial $f(t)\in k[t]$ with no roots in $k$ and homogenize it.  That is, if $f$ has  degree $d$, define $\phi(X,Y)=Y^df(X/Y)$.  You can check that $\phi$ is a homogeneous polynomial of degree $d$ in $X$ and $Y$
in which the coefficient of $X^d$ is nonzero (since that
coefficient comes from the leading coefficient of $f$).  If
$\phi(a,b)=0$ for $a,b\in k$ and $b\neq 0$, then $b^df(a/b)=0$ so $a/b$ is a root of $f$, which is impossible.  On the other hand,
if $b=0$, then because the coefficient of $X^d$ in $\phi$ is nonzero, we must also have $a=0$.  Thus $(0,0)$ is the only zero of $\phi$.

Of course, if one is only interested in the case when $k=\mathbb{R}$, there is in fact no need to reduce to the case when $m=2$: one can just take $$\phi(X_1,\,\ldots\,,X_m)=X_1^2+\,\ldots\,+X_m^2\,,$$ as in Mohan's answer.
