[proofplan] We prove necessity by reducing any primitive integer solution modulo each of $|a|$, $|b|$, and $|c|$; pairwise coprimality forces the remaining coordinate to be invertible modulo the relevant modulus, giving the three quadratic-residue conditions. For sufficiency we use the classical Legendre descent criterion for indefinite primitive diagonal ternary quadratic forms: under the squarefree and pairwise coprime hypotheses, the three residue conditions are exactly the local congruence hypotheses needed to construct a nonzero rational solution. Clearing denominators then gives a nonzero integer solution. [/proofplan]

[step:Reduce any integer solution to a primitive one] Assume first that there is a nonzero triple $(x,y,z) \in \mathbb{Z}^3$ satisfying \begin{align*} ax^2 + by^2 + cz^2 = 0. \end{align*} Let $d = \gcd(x,y,z) \in \mathbb{N}$. Define $(x_0,y_0,z_0) \in \mathbb{Z}^3$ by $x = d x_0$, $y = d y_0$, and $z = d z_0$. Since $(x,y,z) \ne (0,0,0)$, also $(x_0,y_0,z_0) \ne (0,0,0)$, and $\gcd(x_0,y_0,z_0)=1$. Dividing the equation by $d^2$ gives \begin{align*} a x_0^2 + b y_0^2 + c z_0^2 = 0. \end{align*} Thus it is enough, for the necessity direction, to work with a primitive solution. [/step]

[step:Derive the residue condition modulo $|a|$ from a primitive solution]Let $(x,y,z) \in \mathbb{Z}^3$ be a primitive solution. Reducing \begin{align*} ax^2 + by^2 + cz^2 = 0 \end{align*} modulo $|a|$ gives \begin{align*} by^2 + cz^2 \equiv 0 \pmod{|a|}. \end{align*} If $|a|=1$, then the first residue condition is vacuous, since every integer is a quadratic residue modulo $1$. Assume for the rest of this step that $|a|>1$. We claim that $y$ and $z$ are invertible modulo $|a|$. Let $p$ be a prime divisor of $|a|$. Since $a,b,c$ are pairwise coprime, $p \nmid b$ and $p \nmid c$. If $p \mid z$, then the congruence above gives $b y^2 \equiv 0 \pmod p$, hence $p \mid y$. Returning to the original equation modulo $p^2$, the terms $b y^2$ and $c z^2$ are divisible by $p^2$, so $a x^2$ is divisible by $p^2$. Since $a$ is squarefree and $p \mid a$, this forces $p \mid x$. This contradicts $\gcd(x,y,z)=1$. Therefore $p \nmid z$. The same argument with $y$ and $z$ interchanged shows $p \nmid y$. Since every prime divisor of $|a|$ divides neither $z$ nor $c$, both $z$ and $c$ are invertible modulo $|a|$. Multiplying the congruence $by^2 + cz^2 \equiv 0 \pmod{|a|}$ by the inverse of $cz^2$ modulo $|a|$ gives \begin{align*} -bc \equiv (b y z^{-1})^2 \pmod{|a|}, \end{align*} where $z^{-1}$ denotes the inverse of $z$ modulo $|a|$. Hence $-bc$ is a quadratic residue modulo $|a|$.[/step]

[guided]We start from a primitive solution $(x,y,z) \in \mathbb{Z}^3$, meaning $\gcd(x,y,z)=1$, and reduce the equation modulo the coefficient $|a|$. Because the term $a x^2$ is divisible by $|a|$, the equation becomes \begin{align*} by^2 + cz^2 \equiv 0 \pmod{|a|}. \end{align*} If $|a|=1$, the first residue condition is automatic: modulo $1$, every integer is represented by the square $0^2$. Thus there is no inverse to choose and no divisibility argument is needed in that case. We now assume $|a|>1$. To divide by $z^2$ modulo $|a|$, we must prove that $z$ is invertible modulo $|a|$. Let $p$ be any prime divisor of $|a|$. Pairwise coprimality gives $p \nmid b$ and $p \nmid c$. Suppose for contradiction that $p \mid z$. Then the congruence modulo $p$ gives \begin{align*} by^2 \equiv 0 \pmod p. \end{align*} Since $p \nmid b$, this implies $p \mid y$. In the original equation, the terms $b y^2$ and $c z^2$ are then divisible by $p^2$. Therefore $a x^2$ is divisible by $p^2$. Because $a$ is squarefree, the prime $p$ divides $a$ exactly once, so $p^2 \mid a x^2$ forces $p \mid x$. Thus $p$ divides $x$, $y$, and $z$, contradicting primitivity. Hence $p \nmid z$. The same argument with $y$ and $z$ interchanged gives $p \nmid y$. Now $z$ and $c$ are invertible modulo $|a|$. Multiplying \begin{align*} by^2 + cz^2 \equiv 0 \pmod{|a|} \end{align*} by $(cz^2)^{-1}$ yields \begin{align*} -bc \equiv b^2 y^2 z^{-2} = (b y z^{-1})^2 \pmod{|a|}. \end{align*} This writes $-bc$ as a square modulo $|a|$, which is exactly the first residue condition.[/guided]

[step:Derive the residue conditions modulo $|b|$ and $|c|$ explicitly] Reducing the primitive equation modulo $|b|$ gives \begin{align*} ax^2 + cz^2 \equiv 0 \pmod{|b|}. \end{align*} If $|b|=1$, the second residue condition is vacuous. Assume $|b|>1$. Let $p$ be a prime divisor of $|b|$. Since $a,b,c$ are pairwise coprime, $p \nmid a$ and $p \nmid c$. If $p \mid z$, then $a x^2 \equiv 0 \pmod p$, hence $p \mid x$. Returning to the original equation modulo $p^2$, the terms $a x^2$ and $c z^2$ are divisible by $p^2$, so $b y^2$ is divisible by $p^2$. Since $b$ is squarefree and $p \mid b$, this forces $p \mid y$, contradicting $\gcd(x,y,z)=1$. Therefore $p \nmid z$. Interchanging $x$ and $z$ in this argument gives $p \nmid x$. Hence $x$ and $a$ are invertible modulo $|b|$, and multiplying $ax^2 + cz^2 \equiv 0 \pmod{|b|}$ by the inverse of $ax^2$ gives \begin{align*} -ca \equiv (c z x^{-1})^2 \pmod{|b|}. \end{align*} Thus $-ca$ is a quadratic residue modulo $|b|$. Reducing the primitive equation modulo $|c|$ gives \begin{align*} ax^2 + by^2 \equiv 0 \pmod{|c|}. \end{align*} If $|c|=1$, the third residue condition is vacuous. Assume $|c|>1$. Let $p$ be a prime divisor of $|c|$. Since $a,b,c$ are pairwise coprime, $p \nmid a$ and $p \nmid b$. If $p \mid y$, then $a x^2 \equiv 0 \pmod p$, hence $p \mid x$. Returning to the original equation modulo $p^2$, the terms $a x^2$ and $b y^2$ are divisible by $p^2$, so $c z^2$ is divisible by $p^2$. Since $c$ is squarefree and $p \mid c$, this forces $p \mid z$, contradicting $\gcd(x,y,z)=1$. Therefore $p \nmid y$. Interchanging $x$ and $y$ in this argument gives $p \nmid x$. Hence $x$ and $a$ are invertible modulo $|c|$, and multiplying $ax^2 + by^2 \equiv 0 \pmod{|c|}$ by the inverse of $ax^2$ gives \begin{align*} -ab \equiv (b y x^{-1})^2 \pmod{|c|}. \end{align*} Thus $-ab$ is a quadratic residue modulo $|c|$. Together with the first residue condition, every nonzero integer solution implies all three stated residue conditions. [/step]

[step:Invoke Legendre descent to obtain a rational solution from the residue conditions]Assume conversely that the three residue conditions hold. We invoke the independently established [Legendre Descent Criterion for Primitive Diagonal Ternary Forms](/theorems/1178). It states that if $a,b,c \in \mathbb{Z}\setminus\{0\}$ are squarefree and pairwise coprime, if the real quadratic form $Q: \mathbb{Q}^3 \to \mathbb{Q}$ given by \begin{align*} Q(u,v,w) = au^2 + bv^2 + cw^2 \end{align*} is indefinite, and if the three congruence conditions \begin{align*} -bc &\text{ is a quadratic residue modulo } |a|,\\ -ca &\text{ is a quadratic residue modulo } |b|,\\ -ab &\text{ is a quadratic residue modulo } |c| \end{align*} hold, then there exists a nonzero triple $(r,s,t) \in \mathbb{Q}^3$ satisfying \begin{align*} ar^2 + bs^2 + ct^2 = 0. \end{align*} The squarefree, nonzero, and pairwise-coprime hypotheses are exactly those assumed in the theorem. The form $Q$ is indefinite because not all of $a,b,c$ have the same sign, so among the coefficients there is at least one positive and at least one negative coefficient. The three congruence hypotheses are precisely the residue conditions assumed in the converse direction. Therefore the cited descent criterion gives a nonzero rational solution $(r,s,t)$.[/step]

[guided]The sufficiency direction uses an external result, but the result is not the theorem currently being proved; it is the independently established [Legendre Descent Criterion for Primitive Diagonal Ternary Forms](/theorems/1178). We apply that criterion to the quadratic form \begin{align*} Q: \mathbb{Q}^3 &\to \mathbb{Q} \\ (u,v,w) &\mapsto au^2 + bv^2 + cw^2. \end{align*} The criterion requires the following hypotheses. First, the coefficients must be nonzero squarefree integers, which is one of the assumptions of this theorem. Second, the coefficients must be pairwise coprime, also assumed here. Third, the form must be indefinite over $\mathbb{R}$. This follows because the theorem assumes that not all of $a,b,c$ have the same sign; since none of them is zero, at least one coefficient is positive and at least one coefficient is negative, so $Q$ takes both positive and negative values on coordinate vectors. Fourth, the criterion requires exactly the three residue conditions \begin{align*} -bc &\text{ is a quadratic residue modulo } |a|,\\ -ca &\text{ is a quadratic residue modulo } |b|,\\ -ab &\text{ is a quadratic residue modulo } |c|. \end{align*} These are precisely the assumptions in the converse direction. Therefore the descent criterion gives a nonzero rational triple $(r,s,t) \in \mathbb{Q}^3$ such that \begin{align*} ar^2 + bs^2 + ct^2 = 0. \end{align*} This completes the difficult local-to-rational step without using the theorem under proof as an input.[/guided]

[step:Clear denominators to convert the rational solution into an integer solution] Let $(r,s,t) \in \mathbb{Q}^3\setminus\{(0,0,0)\}$ satisfy \begin{align*} ar^2 + bs^2 + ct^2 = 0. \end{align*} Choose $N \in \mathbb{N}$ such that $Nr$, $Ns$, and $Nt$ are all integers. Define \begin{align*} x &= Nr, & y &= Ns, & z &= Nt. \end{align*} Then $(x,y,z) \in \mathbb{Z}^3$ and $(x,y,z) \ne (0,0,0)$. Multiplying the rational equation by $N^2$ gives \begin{align*} ax^2 + by^2 + cz^2 = N^2(ar^2 + bs^2 + ct^2)=0. \end{align*} Hence the original equation has a nonzero integer solution. Combining this with the necessity proved above establishes the claimed equivalence. [/step]