= Solution
For $z\in\mathbb R^n$, define $x_i=z_i^2$. Then $x$ lies in the <nonnegative orthant> and
$$
p(z)=\sum_{i,j}A_{ij}z_i^2z_j^2=x^TAx.
$$
If $A$ is a <copositive matrix>, this is nonnegative for every $z$, so $p$ is a globally <nonnegative polynomial>. Conversely, every $x\geq0$ has the form $x_i=z_i^2$ by taking $z_i=\sqrt{x_i}$. If $p$ is globally nonnegative, then $x^TAx=p(z)\geq0$ for every such $x$, proving that $A$ is a <copositive matrix>. The map $z\mapsto(z_i^2)_i$ covers the entire <nonnegative orthant>, which is the key to both directions.
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