Can Legendre's theorem really help solve this equation: $ax^2+by^2=cz^2$?

Question

let $a,b,c,x,y$ be non-zero positive integers such that $$\gcd(x,y,z)=1$$ find all the non-trivial integral solutions of the diophantine equation:$$ax^2+by^2=cz^2$$ I know that the Legendre's theorem can be helpful in solving this equation but beyond the insight that this theorem provides into the criteria of solvability, it did not help me get any further. Please help.

Answer 1

There is at least one integer solution $(U,V,W)$ to $a U^2 + b V^2 = c W^2$ with small variables, $a U^2 + b V^2 + c W^2 < 4abc.$

Once you have a single such integer solution, all primitive integer solutions are given by taking $\gcd(m,n) = 1,$ then $$ X = -aUm^2 - 2 b V mn + b U n^2, \; \; Y = aVm^2 - 2 a U mn - bV n^2, \; \; Z = (a m^2 + b n^2)W. $$ Then take $$ g = \gcd(X,Y,Z), $$ finally $$ x = X / g, \; \; y = Y/g, \; \; z = Z /g. $$ Depending upon the actual coefficients $a,b,c,$ it may be possible to predict $g,$ resulting in a finite number of slightly different formulas to give all primitive solutions.

Answer 2

We need to write generally speaking the more General equation:

$$aX^2+bXY+cY^2=jZ^2$$

Although I formula solutions recorded, but I see it is of interest expression solutions using any one of the known solution.

If we know what any one solution: $(x,y,z)$ - then you can write a formula for the solutions of this equation.

$$X=jxt^2-cxk^2+2(cyk-jzt)s+(by+ax)s^2$$

$$Y=jyt^2-2jztk+(cy+bx)k^2+2axks-ays^2$$

$$Z=jzt^2-(bx+2cy)kt+czk^2+(bzk-(2ax+by)t)s+azs^2$$

$k,t,s$ - any integer asked us.