How can I show the complete symmetric quadratic form has no zeros? The quadratic complete symmetric homogeneous polynomial in $n$ variables $t_1,\ldots,t_n$ is defined to be
$$h_2(t_1,\ldots,t_n) := \sum_{1 \leq j \leq k \leq n} t_j t_k = \sum_{j=1}^n t_j^2 + \sum_{j<k} t_j t_k.$$
For example, in one variable we have $h_2(t_1) = t_1^2$, in two variables
$$h_2(t_1,t_2) = t_1^2 + t_2^2 + t_1 t_2,$$ and in three 
$$h_2(t_1,t_2,t_3) = t_1^2 + t_2^2 + t_3^2 + t_1 t_2 + t_2 t_3 + t_3 t_1.$$
By a theorem of Jacobi, any quadratic form is conjugate (by an orthogonal transformation) to a diagonal one. It turns out that $h_2(t_1,\ldots,t_n)$ is conjugate to one with one eigenvalue $\frac{n+1}2$ and the others $\frac 1 2$, so there are no nontrivial zeros. 
I came to this question through trying to see if there are any integer solutions. Is there a more clean (perhaps modular) way of seeing there are no integer solutions to $h_2 = 0$?

Anisotropic in $\mathbb Q_2.$ In particular, if
$$  x^2 + y^2 + z^2 + y z + z x + x y \equiv 0 \pmod 4,  $$ then $x,y,z$ are all even. That does it: if there is any integer triple that gives actual $0,$ we can divide through by the gcd of $x,y,z$ to get a primitive representation of $0.$ This little fact about $4$ says that there is no primitive triple  that gives actual $0,$ therefore there is none at all. 
The same argument fails for 4 variables, your $n=4.$ The relevant prime is $5.$ Easy to get divisibility by $5,$ there are $10$ terms, make each $t_i=1.$ It remains easy to get divisibility by $25,$ take your $    t_1=1,   t_2= 2, t_3 =  3, t_4 =   19
$
For an example where real numbers give a different answer from integers, consider $$ w^2 - 2 x^2 + 3 y^2 - 6 z^2.   $$ If $w,x,y,z$ are integers and the form $ w^2 - 2 x^2 + 3 y^2 - 6 z^2   $ is  divisible by $9,$ then $w,x,y,z$ are all divisible by $3.$ So, $ w^2 - 2 x^2 + 3 y^2 - 6 z^2 =0  $ is impossible for nontrivial quadruples in $\mathbb Q_3 $ and in  $\mathbb Z. $
