Indefinite quadratic form universal over negative integers Here's a question that (I hope) may seem very trivial for you, and I hope one of you may provide me with a reference answering it (unless it's a trivial colloquial knowledge). 
Let $f$ be an indefinite ternary quadratic form that is anisotropic (does not represent 0). 
By Dickson's theorem a universal (representing all positive and negative numbers) ternary $f$ is isotropic, and here we see that $f$ cannot be universal. 
Question: can it happen that for any such $f$ there is a \textbf{negative} number $d$ which $f$ does not represent? 
Or: is there an indefinite ternary quadratic form $f$ such that $f$ represents all negative integers, but is not universal (thus avoids some positive number $d$)?
 A: Two parts. If the form is anisotropic, there is a specific prime (actually an even number of primes, by the product formula for the Hilbert Norm Residue symbol) $p$ for which the form is anisotropic. The problem is that the form does not integrally represent anything in the $p$-adic squareclass of the discriminant of the form. This follows from just a few pages in Cassels, rational Quadratic Forms, which I recommend; mostly pages 58-59. 
Next, it is possible to have exceptions that do not directly arise from congruences. The famous 1951 example of Siegel is a genus with two forms,
$$  x^2 - 2 y^2 + 64 z^2, $$
$$  (2x+z)^2 - 2 y^2 + 16 z^2.  $$
Note that the binary $4x^2 + 4 xz + 17 z^2$ is indeed in the same genus as $x^2 + 64 y^2.$ However, while $x^2 + 64 y^2$ is a fourth power in the class group, $4x^2 + 4 xz + 17 z^2$ is a square but not a fourth power.
Now, $(2x+z)^2 - 2 y^2 + 16 z^2$ fails to represent any $m^2,$ where all prime factors of $m$ are $\pm 1 \pmod 8.$ Now, with $  x^2 - 2 y^2 + 64 z^2, $ we can take $x=y=8, z=1$ to get $0.$ Isotropic. At the same time, neither form represents any $\pm 3 \pmod 8.$
A more recent example of Schulze-Pillot and Xu is
$$  x^2 + 100 y^2 - 5 z^2, $$
$$  4 x^2 + 25 y^2 - 5 z^2.  $$
Note that the binary $4x^2  + 25 y^2$ is  in the same genus as $x^2 + 100 y^2.$ However, while $x^2 + 100 y^2$ is a fourth power in the class group, $4x^2 +  25 y^2$ is a square but not a fourth power.
Now, $4 x^2 + 25 y^2 - 5 z^2$ fails to represent any $m^2,$ where all prime factors of $m$ are $\pm 1 \pmod 5.$ Now, with $  x^2 + 100 y^2 - 5 z^2, $ we can take $x=5, y=1, z=5$ to get $0.$ Isotropic. At the same time, neither form represents any $\pm 3 \pmod 5.$
For the moment, I think that we cannot have spinor exceptional integers (as in the two examples) without congruence obstructions, even when all is isotropic. I do not have an immediate proof.  Well, this seems to do it: the genus and the spinor genus coincide unless the discriminant is divisible by $64$ or by $p^3$ for odd prime $p.$ Furthermore, they still coincide unless the diagonalization in $\mathbb Q_p$ is $a x^2 + bp y^2 + c p^2 z^2,$ whereupon we see that half the values $\pmod p$ are not represented, all $n$ with Legendre $(n,p) = -(a,p).$  More messy for $2,$ generally need $\pmod 8.$
A: Representation of a number we write.
$$aX^2+bY^2=cZ^2+q$$
I think that the only way to record the desired polynomial is to use the solutions of any equation.
$$ax^2+by^2=cz^2$$
Knowing the solutions of this equation and substituting them into the linear Diophantine equation.
$$axs+byp-czk=1$$
$(s;p;k) - $ variables which are solutions of this equation.  Then the solution of the first equation can be written as.
$$X=\frac{x}{2}(ck^2-as^2-bp^2+q)+s$$
$$Y=\frac{y}{2}(ck^2-as^2-bp^2+q)+p$$
$$Z=\frac{z}{2}(ck^2-as^2-bp^2+q)+k$$
