RMO 2

Number Theory Level 5

How many integer pairs $(x,y)$ satisfy $x^{2}+ 4y^{2}-2xy-2x-4y-8=0$ ?

The answer is 6.

4 solutions

Brian Charlesworth
Nov 15, 2015

The given equation is that of a rotated ellipse. Noting that $x^{2} - 2xy + 4y^{2} = (x - y)^{2} + 3 y^{2}$ we'll make an attempt at rewriting the equation in the form

$((x - y) - a)^{2} + 3(y - b)^{2} = c$ for some integers $a,b,c.$

Expanding, this becomes $x^{2} - 2xy + y^{2} - 2ax + 2ay + a^{2} + 3y^{2} - 6by + 3b^{2} - c = 0$

$\Longrightarrow x^{2} + 4y^{2} - 2xy - 2ax + (2a - 6b)y + (a^{2} + 3b^{2} - c) = 0.$

Comparing like coefficients between this and the original equation, with the coefficient of $x$ being $-2a = -2$ we set $a = 1.$ Next, this will require that the coefficient of $y$ be such that

$2a - 6b = -4 \Longrightarrow 6b = -4 - 2a = -6 \Longrightarrow b = 1.$

Finally, this will in turn require that

$a^{2} + 3b^{2} - c = 4 - c = -8 \Longrightarrow c = 12.$

So we have factored the original equation to

$(x - y - 1)^{2} + 3(y - 1)^{2} = 12.$

Now since we are looking for integer solutions, both of $(x - y - 1)$ and $(y - 1)$ must be integers. For an equation of the form $A^{2} + 3B^{2} = 12$ to have integer solutions we will require any of

$(A,B) = (0,2), (0,-2), (3, 1), (3, -1), (-3,1), (-3,-1).$

Each of these $6$ pairs results in a distinct solution pair $(x,y).$

For example, with $(A,B) = ((x - y - 1), (y - 1)) = (0,2)$ we find that

$x - y - 1 = 0, y - 1 = 2 \Longrightarrow x = y + 1, y = 3 \Longrightarrow (x,y) = (4,3).$

The other corresponding solutions for pairs $(A,B)$ are $(x,y) = (0,-1), (6,2), (4,0), (0,2)$ and $(-2,0).$

We can thus conclude that there are $\boxed{6}$ ordered integer pairs satisfying the given equation.

Another approach could be considering the following expression as a quadratic equation in x, and then for x to be real, its discriminant should be greater than or equal to 0, which bounds the value of y , ie applying D >= 0 gives -1<= y <= 3.

Hence there are only few values of y, and putting each of them, we can get our desired result.

Harsh Shrivastava - 5 years, 7 months ago

Yes, that method would get the answer more quickly, but I wanted to present a more general approach in case the coefficients of such a problem made the range of possible $y$ values much larger, and to give a sense of the "geometry" of the equation involved.

I could have also written the ellipse equation as

$((x - 2) - (y - 1))^{2} + 3(y - 1)^{2} = 12,$

from which we see that the center of the rotated ellipse lies at $(2,1).$

Brian Charlesworth - 5 years, 7 months ago

You are right that this is the equation of a rotated ellipse centred at $(2,1)$ , but writing the equation as $(x-y-1)^2 + 3(y-1)^2 = 12$ does not show that.

Moving from coordinates $(x,y)$ to coordinates $(x-y-1,y-1)$ is not rotating the system about $(2,1)$ ; it is setting up a set of skew coordinates centred at that point (with the coordinate axes pointing North-South and Northwest-Southeast).

If we rotated the coordinate frame about $(2,1)$ , the equation of the ellipse would end up looking like $aX^2 + bY^2 = c$ , where $X = 2 +dx + ey$ and $Y = 1 + fx + gy$ where $a$ and $b$ are the eigenvalues of the symmetric matrix $\left( \begin{array}{cc} 1 & -1 \\ -1 & 4 \end{array} \right)$ so $a$ and $b$ would be $\tfrac12(5 \pm \sqrt{13})$ , and ${d \choose e}$ and ${f \choose g}$ are the corresponding unit eigenvectors. This is too much work for this problem!

The quickest way to obtain the equation $(x-y-1)^2 + 3(y-1)^2 - 12 \,=\, 0$ is to first complete the square in $x$ , then complete the square in $y$ , with $x^2 - 2xy + 4y^2 - 2x - 4y - 8 \,=\, (x-y-1)^2 + 3y^2 - 6y - 9 \,=\, (x-y-1)^2 + 3(y-1)^2 - 12$

Mark Hennings - 5 years, 7 months ago

@Mark Hennings – I agree with all your comments. I was being a bit informal with the "geometry" of my solution, as my main goal was just to create a sum of squares in whatever way I could so that I could narrow down the integer options for $x,y.$ As you point out, actually finding the equation in standard form for the rotated ellipse would have been more work than necessary for this particular problem. :)

Brian Charlesworth - 5 years, 7 months ago

I too found this approach better.

Rishik Jain - 5 years, 7 months ago

Yes easier approach!

Anik Mandal - 5 years, 6 months ago

You missed the 6th solution which is (4,3).

Kushagra Sahni - 5 years, 6 months ago

That solution is mentioned in my initial example, just before I list the remaining $5$ solutions.

Brian Charlesworth - 5 years, 6 months ago

Chan Lye Lee
Nov 21, 2015

After multiply with 2, the equation $x^{2}+ 4y^{2}-2xy-2x-4y-8=0$ can be written as $(x-2)^2+(x-2y)^2+4(y-1)^2=24$

It is clear now that we need only consider $y=-1, 0, 1, 2, 3$ and finally obtain the six solutions $(x,y)=(0,-1), (4,0), (-2,0), (0, 2), (6, 2), (4,3)$ .

William Isoroku
Nov 19, 2015

Completing the square;

$(x-y)^2-2(x-y)+3y^2-6y=8$

Here's where the magic happens;

$((x-y)^2-2(x-y)+1)+3(y^2-2y+1)=8+1+3=12$

Which becomes;

$(x-y-1)^2+3(y-1)^2=12$

Now here are the conditions; the possible combinations of x and y needs to make the equation as either $3^2+3(1^2)=12$ or $0^2+3(2^2)=12$

Setting up 6 different system of equations with 2 variables and solving for x and y individually will get you the answer.

Took me almost an hour to get the factoring part right :(

Jason Martin
Nov 21, 2015

Treating it as a quadratic polynomial in terms of $x$ , the discriminant is $4(y+1)^2-4(4y^2-4y-8)=12(3-y)(1+y)$ , which is only positive for $y=-1, 0, 1, 2,$ and $3$ . Testing all of these values and using the quadratic formula to see which of these values for $y$ yield integer values for $x$ gives us the $6$ solutions: $(0, -1), (-2, 0), (3, 0), (0, 2), (6, 2), (4, 3)$ .

RMO 2

The answer is 6.

4 solutions

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