$x^2+5xy+3y^2=T$ and $\gcd(x,y)=1$
T is a square number. I need to find which x and y values make the function a square number. Gcd is a greatest common divisor function.
I converted the function Into a pell equation which discriminant is 13, but another way to seach a solution?
in the book Diophantine Equations by Mordell, page 47, Theorem 4 says that the isotropic vectors of a ternary quadratic form can be parametrized with one or more "Pythagorean Triple" type formulas. I wrote one of them below.
Write it as $x^2 + 5xy + 3y^2 = z^2 $ and take
$$ x = u^2 - 2uv - 2 v^2 \; , \; \; y = 2uv + 3 v^2 \; , \; \; z = u^2 + 3 uv - v^2 $$