algebraic geometry and elliptic curves Does $ax^2+by^2=cz^2$ have positive integer solutions?  I know that the solution exists when $(a,b,c)=(1,1,1)$ or $(1,1,n^2+1)$, but I failed to produce a general formula. Any help would be appreciated.
 A: A general solution is due to Legendre, in terms of the Legendre symbol. See for example here: http://public.csusm.edu/aitken_html/notes/legendre.pdf, proposition $2$. See also for Hasse principle in corollary 5, 6 and 7.  
Elliptic curves are not needed, and also no algebraic geometry. Just elementary number theory.
A: As pointed out by D. Burde, there is a general solution by Legendre which gives the necessary conditions of the solvability of,
$$ax^2+by^2+cz^2=0\tag{1}$$
but it is not an instant formula in the sense of,
$$(p^2-nq^2)^2+n(2pq)^2 = (p^2+nq^2)^2\tag{2}$$
which, for $n=1$, gives the Pythagorean triples. You can find a formula like $(2)$ only for particular cases, and not for all general $a,b,c$. 
However, to help you out, if $(1)$ has a non-trivial solution $x, y, z$, then an infinite more can be given as,
$$x_i = x(-a p^2 + b q^2 + c r^2) - 2p(b q y + c r z)$$ 
$$y_i = y(a p^2 - b q^2 + c r^2) - 2q(a p x + c r z)$$
$$z_i = z(a p^2 + b q^2 - c r^2) - 2r(a p x + b q y)$$ 
for arbitrary variables $p,q,r$.
