# Integer Solution of Polynomial (RMO 2015 Chennai Region)

Problem: Solve the equation $$y^3 + 3y^2 + 3y = x^3 + 5x^2 – 19x + 20$$ for positive integers x, y.

Discussion:

$$y^3 + 3y^2 + 3y = x^3 + 5x^2 – 19x + 20$$

Adding 1 to both sides and adjusting we get

$$y^3 + 3y^2 + 3y + 1 = x^3 + 3x^2 + 3x + 1 + 2x^2 – 22x + 20$$

$$\Rightarrow (y+1)^3 = (x+1)^3 + 2x^2 – 22x + 20$$

We have two cases :

Case 1: $$2x^2 – 22x + 20 \le 0$$ implying $$y \le x$$

$$2x^2 – 22x + 20 \le 0$$
$$\Rightarrow x^2 – 11x + 10 \le 0$$
$$\Rightarrow (x-1)(x-10) \le 0$$
$$\Rightarrow 1 \le x \le 10$$

This is an easy check (plug in 1, 2, … , 10).

We find $$x = 1, x = 10;$$ works.

Case 2: $$2x^2 – 22x + 20 \ge 0$$ implying $$y \ge x$$

Then y is at least x+1.

$$(x+1)^3 + 2x^2 – 22x + 20 = (y+1)^3 \ge (x+1+1)^3 = (x+2)^3$$
$$\Rightarrow (x+1)^3 + 2x^2 – 22x + 20 \ge (x+2)^3$$
$$\Rightarrow 2x^2 – 22x + 20 \ge (x+2)^3 – (x+1)^3$$
$$\Rightarrow 2x^2 – 22x + 20 \ge x^3 + 6x^2 + 12x + 1 – x^3 – 3x^2 – 3x – 1$$
$$\Rightarrow 2x^2 – 22x + 20 \ge 3x^2 + 9x -7$$
$$\Rightarrow 0 \ge x^2 + 31x -27$$
But this implies $$\displaystyle{\Rightarrow \frac{-31 – \sqrt{1013}}{2} \le x \le \frac{-31 + \sqrt{1013}}{2}}$$

But x is a positive integer. Hence this is not possible.