Integer Solution of Polynomial | RMO 2015 Chennai Region

Try this problem from RMO 2015 from Chennai Region based on Integer Solution of Polynomial.

Problem: Integer Solution of Polynomial

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

Discussion:

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

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

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

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.

Chatuspathi:

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
CAREERTEAM
support@cheenta.com