Let $q(x,y,z) = ax^2 + by^2 + cz^2$ be a non-singular diagonal ternary quadratic form with integer coefficients. The discriminant $\Delta(q)$ of $q$ is then equal to $abc$, and for any positive number $X$ there are finitely many non-singular diagonal quadratic forms of discriminant at most $X$. Indeed, the number of non-singular diagonal ternary quadratic forms with $|\Delta(q)| \leq X$ is approximated by the integral


$$\displaystyle 8\iiint\_{\substack{1 \leq uvw \leq X \\ u,v,w \geq1}} dudvdw = 8X (\log X)^2 + O(X \log X).$$


A quadratic form $q \in \mathbb{Z}[x,y,z]$ (not necessarily diagonal) is *isotropic* if the equation


$$\displaystyle q(x,y,z) = 0$$


has a solution $(x,y,z) \in \mathbb{Z}^3$. Define


$$\displaystyle N(X) = \#\{q = ax^2 + by^2 + cz^2 : a,b,c \in \mathbb{Z} \setminus \{0\}, |\Delta(q)| \leq X, q \text{ is isotropic}\}.$$


Define 


$$\displaystyle R(X) = \frac{X (\log X)^2}{N(X)}.$$


Is there an asymptotic relation for $R(X)$? Alternatively, how dense are isotropic diagonal ternary quadratic forms among all diagonal ternary quadratic forms?



The density of diagonal isotropic ternary quadratic forms with respect to discriminant
The density must be zero: for each prime $p$, 
the probability that $q$ is *not* $p$-adically isotropic is 
$3/2p + O(1/p^2)$ (when one of $a,b,c$ is a multiple of $p$,
and the product of the other two is $-1$ times a quadratic nonresidue);
thus for any finite set $P$ of primes, the probability is 
$$
\prod\_{p \in P} \left( 1 - \frac3{2p} + O(1/p^2) \right)
= \exp \left[- \sum\_{p \in P} \bigl( 3/(2p) + O(1/p^2) \bigr)\right];
$$
and $\sum\_{p \in P} 3/(2p)$ can be made arbitrarily large, while
$\sum\_{p \in P} O(1/p^2)$ is bounded.


