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 $h_2 = 0$?

Such decisions can be recorded.
Although I like this equation:
$a^2+b^2+c^2+d^2=ab+ac+ad+bc+bd+cd$
And solutions:
$a=p^2-2tps+(3k^2+t^2)s^2$
$b=p^2+3(k+t)^2s^2$
$c=4p^2-4tps+4t^2s^2$
$d=3p^2+6kps+(3k^2+t^2)s^2$
And more:
$a=7p^2+2(3k-t)ps+(3k^2+6kt+7t^2)s^2$
$b=7p^2+6(k-t)ps+3(k-t)^2s^2$
$c=p^2+2(3k+t)ps+(3k+t)^2s^2$
$d=3p^2-6(k+t)ps+(3k^2+6kt+7t^2)s^2$
number $t,k,p,s$  integers and set us.
