Representation of rationals by quadratic form In one paper about number theory author stated 2 lemmas
Lemma 1. If $p$ is a prime $\equiv3(mod $ $4)$ then $x^2+y^2-pz^2$ represents a non-zero rational number $m$ if and only if $m$ is not of the form $kps^2$ with $\left(\frac{k}{p}\right)=1$ or $ks^2$ with $k\equiv p(mod$ $8)$.
Lemma 2. If $p$ and $q$ are odd primes with $p = 1(mod$ $4)$ and $\left(\frac{q}{p}\right)=-1$ then $x^2+qy^2-pz^2$ reperesents a non-zero rational number if and only if $m$ is not of the form $pks^2$ with $\left(\frac{k}{p}\right)=-1$ or $qks^2$ with $\left(\frac{k}{q}\right)=-1$.
To prove these lemmas author refers to general theorem of Hasse-Minkowski. Can they be proved without using such strong results?
 A: Lemma B (for binary) (completing the square and a few cases to check): Given integers, $f(x,y) = a x^2 + b x y + c y^2 ,$ with discriminant $\Delta = b^2 - 4 a c $ not a square. Given a (always positive) prime $r$ with Legendre $(\Delta|r) = -1,$ so that $\Delta \neq 0 \pmod r$ in particular. IF $f(x,y) \equiv 0 \pmod r,$ THEN $x,y \equiv 0 \pmod r$ and  $f(x,y) \equiv 0 \pmod {r^2}.$
It suffices to find out what numbers are represented integrally by these forms. Robinson was very careful about picking these. 
Given $g(x,y,z) = x^2 + y^2 - p z^2,$ with prime $p \equiv 3 \pmod 4.$ If $g(x,y,z) \equiv 0 \pmod 4,$ it follows that $x,y,z$ are all even. We say that $g$ is anisotropic in $\mathbb Q_2;$ this is the most immediate version of anisotropy available for ternary forms. At the same time, if $p \equiv 3 \pmod 8,$ it is easy to check that $g \neq 3 \pmod 8.$ Similarly,  if $p \equiv 7 \pmod 8,$ it is easy to check that $g \neq 7 \pmod 8.$
So, as far as integers go, $x^2 + y^2 - 7 z^2 \neq 4^k (8n+7).$ This is exactly the same behavior as $x^2 + y^2 + z^2$ as regards the prime 2.
Next, $p \equiv 3 \pmod 4,$ so $p$ does not divide $-4$ and $(-4|p)= -1.$ As a result, if $x^2 + y^2 \equiv 0 \pmod p,$ then $x,y \equiv 0 \pmod p$ and $x^2 + y^2 \equiv 0 \pmod {p^2}.$ Thus, if $x^2 + y^2 - p z^2 \equiv 0 \pmod {p^2},$ we then get $x,y,z \equiv 0 \pmod p.$ 
Then what happens at the first power but short of the second? If $p \parallel g, $ which is a popular symbol that means $p|g$ but not $p^2 | g,$ we know that $x,y$ are divisible by $p$ but not $z$. The result is $p$ times $-z^2.$ Now, $p \equiv 3 \pmod 4,$ so $-z^2$ is a quadratic nonresidue; it is not possible to integrally represent $p$ times a quadratic residue $\pmod p.$ So,
$$ x^2 + y^2 - p z^2 \neq 4^k (8n + p); \; \; \;  x^2 + y^2 - p z^2 \neq p^{2k+1} (\mbox{residues} \pmod p); $$
In the expression on the left the $n$ may be postive or negative, and so we can replace $p$ by 3 if $p \equiv 3 \pmod 8,$ or by  7 if $p \equiv 7 \pmod 8.$ Both expression take on a different look if the numbers considered are negative...
Now, Minkowski showed that the form $x^2 + y^2 - p z^2$ integrally represents every other integer. Hasse extended such results to number fields. Leonard Eugene Dickson, in his books, wrote several complete proofs of this more annoying direction for indefinite ternary forms, that they do represent all numbers not ruled out by "conguence considerations." The numbers ruled out are usually referred to as "the progressions" by Dickson, B. W. Jones, and Gordon Pall. And i still encourage you to get the 1950 book of B.W. Jones. As I wrote by email, I do not know the books on this material in Russian. Dickson takes a very elementary approach in Modern Elementary Theory of Numbers (1939). For example, on page 159, exercise 3 is: $x^2 + y^2 - C z^2=0$ has integral solutions, not all $0,$ if and only if $-1$ is a quadratic residue of $C.$   
FOUND IT, no Minkowski or Hasse. Dickson (1939), Theorem 118 on page 164. Let $p$ be an odd prime $p \neq 1 \pmod {24}.$ Then every "T-form" of Hessian $-p$ represents every integer $a$ which is none of the types $4^k(8l+p), p^{2k+1}(pl+R),$ where $k \geq 0$ and $R$ ranges over the $(1/2)(p-1)$ quadratic residues of $p.$ But it never represents an integer of one of those types.  
Dickson proves it the way I or Minkowski would, given "eligible" number $a,$ construct a quadratic form $$ F = a x^2 + b y^2 + c z^2 + 2 r y z + 2 s z x  $$ of the correct determinant (Hessian) and showing that $F$ is integrally equivalent to $x^2 + y^2 - p z^2.$
Alright, Robinson's second form is $x^2 + q y^2 - p z^2,$ with $p \equiv 1 \pmod 4$ and $(q|p) = -1. $ Since $p \equiv 1 \pmod 4,$ it follows that $(-4q|p) = -1.$ By Lemma B above, if $x^2 + q y^2 \equiv 0 \pmod p,$ then $x,y \equiv 0 \pmod p$ and  $x^2 + q y^2 \equiv 0 \pmod {p^2}.$ If $p \parallel x^2 + q y^2 - p z^2,$ the result is $(-p)$ times a residue. But again $p \equiv 1 \pmod 4,$ so this is just $p$ times a quadratic residue. So, the ternary cannot represent $p$ times a NON-residue. The part about $p^2$ applies as before. Next, $(q|q) = -1. $ If $x^2 -p y^2 \equiv 0 \pmod q,$ then $x,y \equiv 0 \pmod q$ and  $x^2 -p y^2 \equiv 0 \pmod {q^2}.$ If $q \parallel x^2 + q y^2 - p z^2,$ the result is $q$ times a residue. So  $q$ times a NON-residue is impossible.
All together, 
$$ x^2 + q y^2 - p z^2 \neq p^{2k+1} (\mbox{NonRes} \pmod p); \; \; \;  x^2 + y^2 - p z^2 \neq q^{2k+1} (\mbox{NonRes} \pmod q). $$
Every other number is so represented. In one case, this is Theorem 119 on page 168 of Dickson 1939. The bad news is that he  restricts the prime we are calling $p,$ because he wants to give a complete list of all such "T-forms." Otherwise everything is there. 
VERY finally, Robinson uses lower case letters to mean integers, while upper case mean rational numbers. Since she chose forms with such strong anisotropy properties, the numbers represented rationally are simply those divided by the squares of integers, as we can replace any $(x,y,z)$ by $(x/m,y/m,z/m).$   
That is probably enough. It seems you want to see Dickson (1939) or a translation. 
