Sunday, November 22, 2015

Find all pairs of integer solutions $x(y^2 + 9) + y(x^2 − 9) + x^2(x − 6) = 0$

Find all pairs $(x,\,y)$ of integers such that they satisfy the equation  $x(y^2 + 9) + y(x^2 − 9) + x^2(x − 6) = 0$

My solution:

Before revealing my method of solving, I wish to tell you how I encountered students kept asking me why should they study quadratic function. How can they be useful.. They said quadratic functions have nothing special, and it's really easy peasy to find for its discriminant, and to completing the square to look for its optimal point and factoring it to investigate its roots.

Yes, that's all that to it for quadratic functions, but when you progress into higher grade, you would encounter problem like solving the equation for integer solutions.

That is when the concept of quadratic function creeps in to assist us in finding all possible integer solutions. How? Continue reading to figure out the answer.

Rewrite the equation as a quadratic in $y$, we have

$x(y^2)+ y(x^2 − 9) + x^2(x − 6) +9x=0$,

its discriminant is hence

$\begin{align*}\text{discriminant}&=(x^2 − 9)^2-4(x)(x^2(x − 6) +9x)\\&=(x^2 − 9)^2-4x^2(x(x − 6) +9)\\&=(x^2 − 9)^2-4x^2(x^2-6x+9)\\&=(x^2 − 9)^2-(2x(x-3))^2\\&=(x^2 − 9)^2-(2x^2-6x)^2\\&=(x^2 -9+2x^2-6x)(x^2-9-2x^2+6x)\\&=(3x^2-6x -9)(-x^2+6x-9)\\&=-(3x^2-6x -9)(x^2-6x+9)\\&=-(x-1)(x+3)(x-3)^2\end{align*}$

We're told $x$ and $y$ are positive integers, hence the discriminant for the quadratic in $y$ must be greater than or equal zero.

Therefore we obtain:

$-(x-1)(x+3)(x-3)^2\ge 0$

Since $(x-3)^2\ge 0$ for all $x\in R$, we must set $-(x-1)(x+3)\ge 0$.

Solving it for the range of $x$ we wind up with $1\le x \le 3$, checking for each case leads us to $(x,\,y)=(-1,\,-4), (2,\,2), (3,\,0)$

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