Solve the equation $3x^2+2y^2-4xy+x-2=0$ in the set of integer numbers.

Question

QUESTION

Solve the equation $3x^2+2y^2-4xy+x-2=0$ in the set of integer numbers.

MY IDEA

I decided to write it as:

$2x^2+2y^2-4xy+x^2+x-2=0={(x\sqrt{2}+y\sqrt{2})}^2+x^2+x-2$

I was thinking of writing $x^2+x-2$ as a perfect square, but I don't know.

Another way might be writing all the equations as a perfect square, which means it should equal $0$.

I'm not that good at this type of problem and I wonder if one of you can explain how you can solve this problem?

Answer 1

You have

$$2(x - y)^2 = 2 - x^2 - x \implies (x - y)^2 = 1 - \dfrac{x(x+1)}{2}.$$

This implies that $1 - x(x+1)/2$ is a perfect square. Since the term $x(x+1)/2$ is always non-negative, then either $(x-y)^2 = 1 - 0 = 1$ or $(x-y)^2 = 1 - 1 = 0$ by means of $x = 0$ or $x = -1$ for the first case, or $x = 1$ or $x = -2$ for the second case. Those values for $x$ yield the solutions $$(x,y) \in \{(-2,-2), (-1,-2), (-1,0), (0,-1), (0,1), (1,1)\}.$$

Answer 2

Solving the quadratic equation $3x^2+(1-4y)x+2y^2-2=0$ in $x$, we obtain the discriminant $$ D=-8y^2-8y+25 $$ which needs to be a square. Hence in particular $D\ge 0$. This is equivalent to $y=1,0,-1,-2$.

Answer 3

$3x^2+2y^2-4xy+x-2=0 $

$2y^2-4xy+3x^2+x-2=0 $

$\begin{array}\\ d &=16x^2-4\cdot 2(3x^2+x-2)\\ &=16x^2-24x^2-8x+16\\ &=-8x^2-8x+16\\ &=-8(x^2+x-2)\\ &=-8(x+2)(x-1)\\ &=-2(4x^2+4x-8)\\ &=-2(4x^2+4x+1-9)\\ &=-2((2x+1)^2-9)\\ \end{array} $

Since we want $d \ge 0$, $(2x+1)^2 \le 9$ so $-3 \le 2x+1 \le 3 $ or $-2 \le x \le 1$.

$d(-2, -1, 0, 1) =(0, 16, 16, 0) $ so the integer values for which $d$ is square are $x=-2,-1, 0, 1$ for which

$\begin{array}\\ y(x=-2, -1, 0, 1) &=\dfrac{4x\pm \sqrt{d(x)}}{4}\\ &=(-2, -1, 0, 1)\pm (0, 1, 1, 0)\\ &=(-2, (-2, 0), (-1, 1), 1)\\ \end{array} $

so $(x, y) =(-2, -2), (-1, -2), (-1, 0), (0, -1), (0, 1), (1, 1) $.

Answer 4

$3x^2 + 2y^2 - 4xy + x - 2 = 2(x - y)^2 + (x + 2)(x - 1) = 0$

$2(x - y)^2 = 0$ AND $(x + 2)(x - 1) = 0$
$x = y$ AND $x = -2$ or $x = 1$
So, $ y = -2$ or $y = 1$

For the other solutions, solve: $2(x - y)^2 = - (x + 2)(x - 1)$

So,

1. $x < -2$ AND $x > 1$ Impossible!!
2. $x > -2$ AND $x < 1$ So, $x = -1$ or $x = 0$

Solving for $y$, we get $y = -1$ (for $x = -1$) and we get $y = 0$ (for $x = 0$)

The solution set is $\{-2, -1, 0, 1\}$