I'm trying to find all anisotropic quadratic forms over $F_p=Z/pZ$. I have found that "If $F$ is a finite field and $(V, q)$ is a quadratic space of dimension at least three, then it is isotropic" (here https://en.wikipedia.org/wiki/Isotropic_quadratic_form). So I wonder if the fact is true and if it is, why. Thank you
Quadratic forms over $F_p$ in more than $2$ variables are isotropic
finite-fieldsgeometryquadratic-forms
Related Solutions
A precise definition is given in the wikipedia link on (non-degenerate) quadratic forms. Since both your fields $\mathbb{Q}$ and $\mathbb{Q}_p$ have characteristic zero, you can always write $f(x)=x^TAx$ with a symmetric matrix $A$, which can be even transformed into a diagonal matrix $D$. Indeed, the last step fails for fields of characteristic $2$, but this is not your concern. Then $f$ is non-degenerate if and only if all diagonal elements of $D$ are non-zero.
ADDED: it all works. In Dickson's METN, we find, theorem 117 on page 164, that $w^2 + u^2 - 6 z^2$ is the only equivalence class of "determinant" $-6,$ attributed to Ross. Then on page 170, exercise 2, we see that $w^2 + u^2 - 6 z^2$ integrally represents all numbers other than $$ 4^k \left( 16 m + 6 \right) \; , \; \; 9^k \left( 9 m + 6 \right) \; . $$ In particular, $9-6 = 3,$ and all other primes $p \equiv 3 \pmod 8$ are so represented.
ORIGINAL:
This is provisional, I will think about it overnight. The form $$ w^2 - 2 y^2 + 3 y^2 - 6 z^2 $$ is anisotropic in $\mathbb Q_3$ three-adics. It is elementary to show that the form (integrally) represents the product of any two integers it integrally represents, that it does represent $-1,$ and that it integrally represents all (positive) primes $$ p \equiv 1,5,7 \pmod 8 $$
A computer search strongly suggests that it also represents all primes $p \equiv 3 \pmod 8.$ If that can be proved, we are done. Note tht the output below has $x=y,$ meaning that we are looking at an indefinite ternary $w^2 + v^2 - 6 z^2.$
Mon Oct 28 11:10:15 PDT 2019
prime p == 3 mod 8, and demanding x = y:
p w x = y z
3 0 3 3 1
11 1 4 4 1
19 0 5 5 1
43 0 7 7 1
59 1 8 8 1
67 0 11 11 3
83 4 11 11 3
107 1 16 16 5
131 4 11 11 1
139 0 17 17 5
163 0 13 13 1
179 4 13 13 1
211 0 19 19 5
227 4 19 19 5
251 1 16 16 1
283 0 17 17 1
307 0 19 19 3
331 0 25 25 7
347 1 20 20 3
379 0 23 23 5
419 5 20 20 1
443 7 20 20 1
467 11 20 20 3
491 1 28 28 7
499 0 35 35 11
523 0 23 23 1
547 0 29 29 7
563 4 29 29 7
571 0 25 25 3
587 1 40 40 13
619 0 25 25 1
643 0 37 37 11
659 4 37 37 11
683 4 31 31 7
691 0 29 29 5
739 0 35 35 9
787 0 29 29 3
811 0 31 31 5
827 4 31 31 5
859 0 47 47 15
883 0 37 37 9
907 0 31 31 3
947 4 35 35 7
971 1 32 32 3
1019 1 32 32 1
1051 0 49 49 15
1091 4 35 35 5
1123 0 43 43 11
1163 7 40 40 9
1171 0 35 35 3
1187 4 35 35 3
1259 7 44 44 11
1283 8 35 35 1
1291 0 55 55 17
1307 1 40 40 7
1427 8 37 37 1
1451 1 40 40 5
1459 0 53 53 15
1483 0 47 47 11
1499 4 47 47 11
1523 8 53 53 15
1531 0 41 41 5
1571 4 43 43 7
1579 1 48 48 11
1619 5 40 40 1
1627 0 41 41 3
1667 5 44 44 7
1699 0 43 43 5
1723 0 47 47 9
1747 4 45 45 7
1787 1 44 44 5
1811 4 43 43 3
1867 7 48 48 9
1907 5 44 44 3
1931 1 44 44 1
1979 1 52 52 11
1987 3 52 52 11
2003 5 52 52 11
2011 0 55 55 13
2027 4 55 55 13
2083 0 53 53 11
2099 4 53 53 11
2131 4 51 51 9
p w x = y z
Best Answer
First, an observation: if the number of variables of $q$ is less than the dimension, then we can get an isotropic vector by assigning $0$ to the appearing variables and $1$ to the non appearing ones. So we now assume the number of variables $n$ to be the dimension.
Then your result is an specific instance of the Chevalley-Warning theorem: If a set of polynomials $\{f_i\}$ over a finite field satisfies $n>\sum_i d_i$, where $n$ is the number of variables and $d_i$ the degree of $f_i$, then the system of equations $\{f_i=0\}$ has a number of solutions which is a multiple of the characteristic of the finite field.
In your case, you have just one polynomial $q$ of degree $2$ and $n\geq 3>2$, so the number of solutions is a multiple of $p$, hence can be $0,p,2p\ldots$. But we already know that $(0,\ldots,0)$ is a solution, so there must be at least other $p-1$ nonzero solutions $v$ which give $q(v)=0$.