How can I show the complete symmetric quadratic form has no zeros?

Question

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 $h_2 = 0$?

Answer 1

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.

Answer 2

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. $

Answer 3

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.