Find the integer solution $x^2+3x+9=9y^2$.............(1)

This is a beautiful quadratic Diophantine's equation that requires an integer solution. So how do we go from here? 😁😁😁

I'm going to try as much as possible to make the solution easily comprehensible for those who might find it a bit confusing or difficult.

Looking at the equation closely, you will realize the left hand side is just our normal everyday quadratic equation which can easily be factored. The right hand side is also a quadratic, but the problem is, the whole equation has two variables.

We are going to solve this problem by treating the whole equation as a single quadratic equation, so move $9y^2$ to the left hand side so that you can equate the equation to zero.
$x^2+3x+9-9y^2=0$
Use quadratic formula to find the factors,
Let $a=1,b=3,c=9-9y^2$
$x=\frac{-3\pm\sqrt{9-4(9-9y^2)}}{2}$
$x=\frac{-3\pm\sqrt{9-36+36y^2}}{2}$
$x=\frac{-3\pm\sqrt{36y^2-27}}{2}$
Factor our $9$
$x=\frac{-3\pm\sqrt{9(4y^2-3)}}{2}$
$x=\frac{-3\pm{3}\sqrt{(4y^2-3)}}{2}$......................(2)
Now that we have arrived at this, what do we do next? This is where it gets confusing and boring 😔🤔. 
All we need to do is some algebraic manipulations, so your method might be different from mine, but what matters is the goal(integers).

What I'm going to do now is,
Let $\sqrt{4y^2-3}=k$ so that $4y^2-3=k^2$...............(3)
Therefore, $x=\frac{-3\pm{3k}}{2}$.
Rearrange (3) so it forms an equation with difference of two squares.
$4y^2-k^2=3$
$(2y-k)(2y+k)=3$
All we need to do now is find integers for which we will equate the factors to find $y$.
We will equate each of the factors to $(1,3,-1,-3)$.

So equate $2y-k=1$ and $2y+k=3$
Sloving simultaneously for $y$ yields 👇
$y=1$
Also, equate $2y-k=-1$ and $2y+k=-3$ and solve simultaneously for $y$ to get 👇
$y=-1$
Now there are two values of $y=1,-1$ we will plug in these values of $y$ into equation to obtain the integer values of $x$.
👉For $y=1$
$x^2+3x+9=9(1)^2$
$x^2+3x+9-9=0$
$x^2+3x=0$
Solve quadratically to get 👇
$x=0,-3$

👉 For $y=-1$
$x^2+3x+9=9(-1)^2$
$x^2+3x+9-9=0$
$x^2+3x=0$
Solve quadratically to get 👇
$x=0,-3$
Hence, we conclude the integer solutions for the equation for both $x$ and $y$ are 👇
$x=0,-3$ and $y=1,-1$
And that's it. 😁😁😁😁😂😁😁😁😁
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