# Why is considering only quadratic in one of the variables of a two variable quadratic sufficient for calculating roots

Find the $$positive$$ integral solutions to $$7x^2-2xy+3y^2-27=0$$

My solution:

Assuming the quadratic in $$x$$ , if we assume one root to be integral , the other has to be rational (as y must be an integer to satisfy the condition so ,product of roots is rational)

For the roots to be rational the $$discriminant$$ has to be a perfect square. We get the discriminant($$\Delta$$) as

$$\Delta=4(189-20y^2)$$ which has to be a perfect square.

So we get $$y^2$$=$$1,9$$ hence $$y$$ as $$1,3$$

putting the values back we get the pair $$(x,y)=(2,1)$$

If we again make a quadratic in $$y$$ we get the same solution. Hence considering a quadratic in $$x$$ only is sufficient.

My Question:

I don't get the fact or intuition behind as to why does considering the quadratic in either $$x$$ or $$y$$ is self sufficient as it is not symmetric. If someone could provide me the intuition or proof behind as to why both of them lead to the same results , it would be of great help.

Note: All the similar questions I have encountered can be solved by considering only quadratic in either $$x$$ or $$y$$ only, so i assume it is general.

Thanks.

You are looking for solutions among $$(x,y)$$ where $$x$$ is restricted to be integer. But clearly, the integer solutions in $$y$$ all belong to this subset and you won't miss any.

Assuming $$a$$ and $$b$$ are nonzero, we can solve the equation$$ax^2+bxy+cy^2+d=0 \tag{1}\label{1}$$for $$x$$ by considering the equation as a quadratic in $$x$$, namely$$x=\frac{-by \pm \sqrt{(by)^2-4a(cy^2+d)}}{2a}, \tag{2}\label{2}$$and for $$y$$ by considering the equation as a quadratic in $$y$$, namely$$y=\frac{-bx \pm \sqrt{(bx)^2-4c(ax^2+d)}}{2c}.\tag{3}\label{3}$$

Please note that \ref{1}, \ref{2}, and \ref{3} are the same equations, so solving each of them alone is sufficient to find all roots.

So, for example, if you want to find all positive integer roots of the equation$$7x^2-2xy+3y^2-27=0,$$both \ref{2} and \ref{3} will give you the same results.

• Thanks for your answer, if i had been allowed i would have accepted yours also as – Satwik Jul 19 '20 at 13:53

Assume that someone told us that $$y=1$$, then the question would be to find the integer solution to: $$7x^2-2x+3-27=0.$$

In other words the intuition comes from the fact that if $$(x_0,y_0)$$ is a root of $$F(x,y)=0$$, then $$x_0$$ is a root of $$F(x, y_0)=0$$. Which is easy to prove by contradiction.

• I still could not get why considering only one of them is sufficient and if we dont know the integral value that satisfies the equation, if you would please elaborate – Satwik Jun 3 '20 at 16:57