Solving a Quadratic Ternary Form with Large Coefficients I encountered this expression 
$$1215696x^2+566544y^2-103776z^2=0$$
which I've understood is called a "quadratic ternary form" and have been trying to find solutions $x,y,z\in\mathbb{Z}$
One can firstly remove a common factor of $48$ and obtain
$$25327x^2+11803y^2-2162z^2=0$$
At this point, the coefficients are already square free. I'm trying to follow Legendre's theorem on these forms (which I understand to characterize solvability) and Lagrange's method of descent (which I understand finds initial solutions).  I've had no luck so far.
I've noted too that each of the coefficients are congruent numbers. Specifically, they are of the form $4mn(m^2-n^2)$ for $m,n\in\mathbb{Z}$. Would this fact be enough to determine solvability?
 A: This is impossible in the $23$-adic numbers. The calculation is that, if
$$ 25327 x^2 + 11803 y^2 \equiv 0 \pmod {23}, $$ then, in fact, both $x$ and $y$ are divisible by $23.$ Therefore $$ 25327 x^2 + 11803 y^2 \equiv 0 \pmod {23^2}, $$ from which we get $z \equiv 0 \pmod {23.}$ This contradicts the standard assumption that we have a solution with $\gcd(x,y,z) = 1.$ 
Calculations: $2162 = 2 \cdot 23 \cdot 47. \; \; \; $  $25327 = 4 + 23 \cdot 1101. \; \; \; $ $11803 = 4 + 23 \cdot 513.$ 
$$ 4 x^2 + 4 y^2 \equiv 0 \pmod {23}. $$
$$  x^2 +  y^2 \equiv 0 \pmod {23}. $$
ASSUME that $y$ is not divisible by $23.$ Then we can do
$$  x^2   \equiv - y^2 \pmod {23}. $$
$$  \frac{x^2}{y^2}   \equiv - 1 \pmod {23}, $$
$$  \left(\frac{x}{y} \right)^2  \equiv - 1 \pmod {23}. $$
This contradicts the fact that $-1$ is not a quadratic residue mod any prime $q \equiv 3 \pmod 4,$ which includes $q=23.$
In case of interest: by the product formula for the Hilbert Norm Residue symbol, there must be an even count of primes  which block your problem. There is no problem with the prime $2,$ but there are several others as factors. 
NOTE: I like the treatment in Cassels, Rational Quadratic Forms. He makes it very clear that one must check each prime separately, and in earlier pages shows ways to accomplish that.  
