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 $h_2 = 0$?

Well there is the trivial solution $\tilde{t}=(t_1,t_2,\cdots ,t_n)=(0,0,\cdots ,0)$. Now write $$\displaystyle h_2(\tilde{t})=\frac{1}{2}\left(\sum_{i=1}^{n}t_i^2\right)+\frac{1}{2}\left(\sum_{i=1}^{n}t_i\right)^2\ge 0$$ where equality only occurs when everything is zero. I suppose I understood the question correctly. 
