Determine the pairs of integers $(x,y)$ that verify the relation: $x^2y^2+2xy+36=3y^2+8x^2$

Question

the question

Determine the pairs of integers $(x,y)$ that verify the relation:

$$x^2y^2+2xy+36=3y^2+8x^2$$

the idea

Fist of all I tried getting everything on the LHS and write it as a product of numbers or a sum of perfect squares that equal 0, but got to nothing useful.

Then I tried writing everything as $(xy+1)^2=3y^2+8x^2-35$ or as $(x^2-2)(y^2-7)+22=(x-y)^2$, but again you can see we can do nothing with these.

Hope one of you can help me! Thank you!

Answer 1

A solution based on different ideas from the comments. Let us solve this equation for $y$, we get $$y=\frac{-x+ \sqrt{8x^4-59x^2+108}}{x^2-3},\tag1$$ $$y=\frac{-x- \sqrt{8x^4-59x^2+108}}{x^2-3}.\tag2$$

Let us prove that if $|x|>3$ then $y$ in the form $(1)$ is in $(2,3)$: $$2<\frac{-x+ \sqrt{8x^4-59x^2+108}}{x^2-3}<3.$$

Multiply by $x^2-3>0$: $$2x^2-6+x<\sqrt{8x^4-59x^2+108}<3x^2-9+x.$$

Square all sides (they are positive while $|x|>3$): $$4x^4+36+x^2-24x^2-12x+4x^3<8x^4-59x^2+108<$$ $$<9x^4+81+x^2-54x^2+6x^3-18x.$$

Left inequality is: $$4x^4-4x^3-36x^2+12x+72>0$$ $$\iff x^4-x^3-9x^2+3x+18>0$$ $$\iff (x+2)(x-3)(x^2-3)>0,$$

which is true for $|x|>3$. The right inequality is:

$$x^4+6x^3+6x^2-18x-27>0$$ $$\iff (x+3)^2(x^2-3)>0,$$

which is again true for $|x|>3$.

Now prove the similar statement for $y$ in the form $(2)$: $$|x|>3\implies -2>\frac{-x- \sqrt{8x^4-59x^2+108}}{x^2-3}>-3.$$

Multiply by $x^2-3>0$: $$-2x^2+6+x>-\sqrt{8x^4-59x^2+108}>-3x^2+9+x$$

$$\iff 2x^2-6-x<\sqrt{8x^4-59x^2+108}<3x^2-9+x$$

Square all sides (they are positive while $|x|>3$): $$4x^4+36+x^2-24x^2+12x-4x^3<8x^4-59x^2+108<$$ $$<9x^4+81+x^2-54x^2-6x^3+18x.$$

Left inequality is: $$4x^4+4x^3-36x^2-12x+72>0$$ $$\iff x^4+x^3-9x^2-3x+18>0$$ $$\iff (x-2)(x+3)(x^2-3)>0,$$

which is true for $|x|>3$. The right inequality is:

$$x^4-6x^3+6x^2+18x-27>0$$ $$\iff (x-3)^2(x^2-3)>0,$$

which is again true for $|x|>3$.

Now we could do the same thing solving the initial equation for $x$ and repeating the similar reasoning showing that $|y|$ can’t be more than $2\sqrt2$. Or we can just solve seven linear/quadratic equations putting $x=-3,-2,-1,0,1,2,3$ in the initial equation and getting all the solutions: $(-3,-2),(-3,3),(-2,2),(2,-2),(3,-3),(3,2)$.

Answer 2

Grouping the terms of $y$ gives $$(x^2-3)y^2 + (2x)y + (36-8x^2).$$ The quadratic formula implies (just as @Dan suggested) that $$y = \frac{-x\pm \sqrt{8x^4-59x^2+108}}{x^2-3}.$$ Since we require $y$ to be integer we need $8x^4-59x^2+108$ to be a square.

Substituting $u=x^2$ we can look for $u$ such that $8u^2-59u+108$ is a square. The Diophantine equation $8u^2-59u+109=w^2$ for some integer $w$ admits the following recurrence relation for $u$: $u_0 = 4$, $u_1=9$, $u_2=184$, $u_n=u_{n-3}-35u_{n-2}+35u_{n-1}$. Where $u_0, u_1, \dots$ represent all possible solutions for $u$ to the Diophantine equation.

Using that recurrence we can easily see that any $x$ satisfying the original equation must satisfy $x=\pm 2$, $x=\pm 3$, or $|x| > 10^3$ (note that you must check all solutions of $u$ up to $10^6$ to verify this, luckily there are only $5$ of these).

If $|x| > 10^3$ then $|y|<3$ (due to the comment of @sbares, we know that this bound holds as $x$ get large).

We thus have $x = \pm 2$, $x = \pm 3$, or $y\in \{ 0, \pm 1, \pm 2 \}$. Substituting these values in the original equation gives the solutions $(x, y) \in {(−3,−2),(−3,3),(−2,2),(2,−2),(3,−3),(3,2)}$. (As @Dan computed earlier on).