Question on quadratic forms I know a theorem which says: 

If a non-singular quadratic form (homogeneous polynomials of degree $2$) over a field $K$ represents zero non-trivially (i.e., there is a nontrivial solution of the equation $F=0$), then it represents all the non-zero elements of field $K$. 

A quadratic form $F$ is said to represent $a$ (in $K$) if there exists $x_1,x_2,\dots,x_n$ such that $F(x_1,x_2,\dots,x_n)= a$.  

My question is whether the converse of the theorem is true when char $K=0$.

When $K$ is a finite field of order $3$, then $X^2+Y^2$ contradicts the converse of the theorem. 

field is $\mathbb Q,$ form is $$ w^2 - 2 x^2 + 3 y^2 - 6 z^2.   $$
Background. This is Theorem 19 on page 53 of Burton W. Jones, The Arithmetic Theory of Quadratic Forms. Somewhat better tables and instructions for computations in J.W.S. Cassels, Rational Quadratic Forms, pages 55-60, tables pages 43-44.
As it happens, this form is universal over the integers with integer variable values, which is an extra result over the Hasse-Minkowski stuff over the rationals.
Lots of detail required for all claims except one: the form is anisotropic in $\mathbb Q$ because it is anisotropic in the 3-adic numbers, and it suffices to consider rational integers. 
EXERCISE. If
$$  w^2 - 2 x^2 + 3 y^2 - 6 z^2 \equiv 0 \pmod 9,  $$
then $$ w,x,y,z  \equiv 0 \pmod 3. $$
EEDDDIIITTTT, June 26.   The fact that, if $w^2 - 2 x^2 + 3 y^2 - 6 z^2$ integrally represents two numbers, it also integrally represents their product, is an exercise in quaternion multiplication (with evident coefficients). The fact that $w^2 - 2 x^2 + 3 y^2$ integrally represents all positive primes follows from the evident sections $w^2 + y^2 $ when $x=y,$ and $w^2 - 2 x^2$ when $z=0.$ For the primes $q\equiv 3 \pmod 8,$ we do need material from Dickson's 1939 book, where he prefers to write it as $$ w^2 - 2 (-x-3y)^2 + 3 (y+2z)^2 = w^2 + x^2 - 6 y^2.  $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
