How to find all rational solutions of $\ x^2 + 3y^2 = 7 $?
The key here is to understand where the substitution $$m = \frac{y}{x + 1}$$ comes from. Geometrically, this variable represents the slope of the line between a point $(x, y)$ and the point $(-1, 0)$. What you are then doing is considering a rational slope $m$, taking the line $L$ through $(-1,0)$ of slope $m$, and solving for the second intersection point of this line with the conic $x^2+y^2=1$ (the first intersection point being $(-1,0)$).
This same approach works for any conic, as long as you have a single rational point on the conic to play the role of $(-1,0)$. In the case of $x^2+3y^2=7$, for instance, you could take $(2,1)$ as your initial point. So then you would define $$m = \frac{y - 1}{x - 2}$$ and solve for $(x,y)$ as the second point where the line $L$ through $(2,1)$ of slope $m$ intersects the conic $x^2+3y^2=7$.
Using the method of pg 7 of this paper on this related equation $$x^2+3y^2=7z^2 \quad \text{with initial solution} \quad (x,y,z)=(2,1,1)$$
A line $y=t(x-2)+1$, which will cut through the ellipse $x^2 + 3y^2 = 7$ at rational points if $t$ is rational.... when substituted into the ellipse yields:
$$\begin{align} x^2+3\left[t(x-2)+1\right]^2&=7 \\ x^2+3\left[t^2(x^2-4x+4)+2t(x-2)+1\right]&=7 \\ (1+3t^2)x^2+(-12t^2+6t)x+(12-12t+3-7)&=0 \\ \text{vieta: the two roots,} \quad x_1, x_2 \quad \text{are such that} \quad -(x_1+x_2)&=\frac{-12t^2+6t}{1+3t^2} \\ \text{since} \quad x_1=2, \quad \text{we have} \quad x_2(t)=\frac{12t^2-6t}{1+3t^2}-2 &=\frac{6t^2-6t-2}{1+3t^2} \\ \text{substituting this} \quad x(t) \quad \text {into the line:} \quad y&=t\left(\frac{6t^2-6t-2}{1+3t^2}-2\right)+1 \\ y(t)=t\left(\frac{-6t-4}{1+3t^2}\right) + \frac{1+3t^2}{1+3t^2}&=\frac{-3t^2-4t+1}{1+3t^2} \end{align}$$
Letting $y(t)\to |y(t)|$, Solution set with one parameter:
$$\begin{cases} x(t)&=\frac{6t^2-6t-2}{1+3t^2} \\ y(t)&=\frac{3t^2+4t-1}{1+3t^2} \end{cases}$$
$t$ was rational, so let $t=\frac{m}{n}$
Solution set now with two parameters: $$\begin{align} x(m,n)&=\frac{6m^2-6mn-2n^2}{n^2+3m^2} \\ y(m,n)&=\frac{3m^2+4mn-n^2}{n^2+3m^2} \end{align}$$
Solving a Diophantine equation of the form $x^2 = ay^2 + byz + cz^2$ with the constants $a, b, c$ given and $x,y,z$ positive integers
The problem is solved definitely. Why is the question constantly recurs to me is not clear. Consider two options to solve this problem. The first option is to directly solve the equation without knowing whether there are solutions.
$$aX^2+bXY+cY^2=jZ^2$$
Solutions can be written if even a single root. $\sqrt{j(a+b+c)}$ , $\sqrt{b^2 + 4a(j-c)}$ , $\sqrt{b^2+4c(j-a)}$
Then the solution can be written.
$$X=(2j(b+2c)^2-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+$$
$$+2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+(j\mp \sqrt{j(a+b+c)})p^2$$
$$Y=(2j(2j-b-2a)(b+2c)-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+$$
$$+2((2j-2a-b)\sqrt{j(a+b+c)}\mp{j(b+2c)})sp+(j\mp\sqrt{j(a+b+c)})p^2$$
$$Z=(2j(b+2c)^2-(b^2+4c(j-a))(a+b+c\pm\sqrt{j(a+b+c)}))s^2+$$
$$+2(b+2c) ( \sqrt{j(a+b+c)} \mp{j})sp + ( a + b + c \mp \sqrt{j(a+b+c)})p^2$$
In the case when the root $\sqrt{b^2+4c(j-a)}$ whole. Solutions have the form.
$$X=((2j-b-2c)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+$$
$$+2(4ac+b(2j-b)\pm{(2j-b-2c)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Y=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(2j-b-2a\mp\sqrt{b^2+4c(j-a)}))s^2+$$
$$+2(4ac+b(2j-b)\pm{(b+2a)}\sqrt{b^2+4c(j-a)})sp+(2j-b-2a\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Z=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+$$
$$+2(4ac+b(2j-b)\pm {(b+2a)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
In the case when the root $\sqrt{b^2+4a(j-c)}$ whole. Solutions have the form.
$$X=(2j^2(b+2a)-j(a+b+c)(2j-2c-b\pm\sqrt{b^2+4a(j-c)}))p^2+$$
$$+2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(2j-2c-b\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Y=(2j^2(b+2a)-j(a+b+c)(b+2a\pm\sqrt{b^2+4a(j-c)}))p^2+$$
$$+2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Z=j(a+b+c)(b+2a\mp\sqrt{b^2+4a(j-c)})p^2+$$
$$+2((a+b+c)\sqrt{b^2+4a(j-c)}\mp{j(b+2a)})ps+ (b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
Since these formulas are written in general terms, require a certain specificity calculations.If, after a permutation of the coefficients, no root is not an integer. You need to check whether there is an equivalent quadratic form in which, at least one root of a whole. Is usually sufficient to make the substitution $X\longrightarrow{X+kY}$ or more $Y\longrightarrow{Y+kX}$ In fact, this reduces to determining the existence of solutions in certain Pell's equation. Of course with such an idea can solve more complex equations. If I will not disturb anybody, slowly formula will draw. number $p,s$ integers and set us. I understand that these formulas do not like. And when they draw - or try to ignore or delete. Formulas but there are no bad or good. They either are or they are not.
In equation $$aX^2+bY^2+cZ^2=qXY+dXZ+tYZ$$
$a,b,c,q,d,t - $ integer coefficients which specify the conditions of the problem. For a more compact notation, we introduce a replacement.
$$k=(q+t)^2-4b(a+c-d)$$
$$j=(d+t)^2-4c(a+b-q)$$
$$n=t(2a-t-d-q)+(2b-q)(2c-d)$$
Then the formula in the general form is:
$$X=(2n(2c-d-t)+j(q+t-2b\pm\sqrt{k}))p^2+$$
$$+2((d+t-2c)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2$$
$$Y=(2n(2c-d-t)+j(2(a+c-d)-q-t\pm\sqrt{k}))p^2+$$
$$+2((d+t-2c)\sqrt{k}\mp{ n })ps+(q+t+2(d-a-c)\pm\sqrt{k})s^2$$
$$Z=(j(q+t-2b\pm\sqrt{k})-2n(2(a+b-q)-d-t))p^2+$$
$$+2((2(a+b-q)-d-t)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2$$
And more.
$$X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+$$
$$+2((2b-q-t)\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2$$
$$Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2+$$
$$+2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2$$
$$Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+$$
$$+2((2b-q-t)\sqrt{j}\mp{n})ps+(2(a+b-q)-d-t\pm\sqrt{j})s^2$$
$p,s - $ are integers and are given us. Since formulas are written in general terms, in the case where neither the root is not an integer, it is necessary to check whether there is such an equivalent quadratic form in which at least one root of a whole. If not, then the solution in integers of the equation have not.
Let's use this formula to solve this equation. For example use 2 formula.
$$aX^2+bXY+cY^2=jZ^2$$
$a=1 ; b= 0; c=3; j=7$
$$X^2+3Y^2=7Z^2$$
$\sqrt{0^2+4*1*(7-3)}=4$
$$X=-70p^2+14ps+2s^2$$
$$Y=14p^2+14ps-s^2$$
$$Z=-28p^2+2ps-s^2$$
$$***$$
$$X=2(7p^2+7ps+s^2)$$
$$Y=42p^2+14ps+s^2$$
$$Z=28p^2+10ps+s^2$$
When considering a different root.
$$3X^2+Y^2=7Z^2$$
$a=3 ; b=0 ; c=1; j=7$
Use 3 formula.
$$X=18s^2+10ps+p^2$$
$$Y=-6s^2+6ps+2p^2$$
$$Z=12s^2+6ps+p^2$$
$$***$$
$$X=2s^2+6ps+p^2$$
$$Y=-10s^2-2ps+2p^2$$
$$Z=4s^2+2ps+p^2$$