TITLE: polynomials in many variables and Hasse principle
QUESTION [5 upvotes]: I was wondering whether there exists any result of the form 

"if $f \in \mathbb{Z}[x_1, ..., x_k]$ is a polynomial (not form! I don't require homogeneity) of total degree $n$, with $k \geq \delta n$ for some explicit $\delta$, then $f(x) = 0$ satisfies the Hasse principle."

I'm particularly interested in the case $n = 4$. Thanks!
EDIT: I'm interested in integral solutions.

REPLY [10 votes]: The integral Hasse principle can fail for polynomials of degree $2$ (or $4$), even if the number of variables is huge!  For example, if $k \equiv 1 \pmod{8}$, then
$$(2x_1-1)^2 + \cdots + (2x_k-1)^2 = 1$$
has the solution $(1,1/2,\ldots,1/2)$ over $\mathbf{R}$ and $\mathbf{Z}_p$ for all odd $p$, and has a solution $(a,1,\ldots,1)$ over $\mathbf{Z}_2$ for some $a$ near $1$ by Hensel's lemma, but it has no solution over $\mathbf{Z}$ for $k>1$ since each square on the left is at least $1$.
For a much more detailed study of the integral Hasse principle, see Colliot-Thélène and Xu, 
Brauer-Manin obstruction for integral points of homogeneous spaces and representation by integral quadratic forms, Compos. Math. 145 (2009), no. 2, 309–363.