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

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


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

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 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.

  • $\begingroup$ Thank you! This is a satisfactory answer to my question. $\endgroup$ – Stanley Yao Xiao May 1 '17 at 13:22

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