Try this problem from RMO 2015 from Chennai Region based on Integer Solution of Polynomial.
Problem: Integer Solution of Polynomial
Solve the equation $latex y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20 &s=2 $ for positive integers x, y.
Discussion:
$latex y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20 &s=2 $
Adding 1 to both sides and adjusting we get
$latex y^3 + 3y^2 + 3y + 1 = x^3 + 3x^2 + 3x + 1 + 2x^2 - 22x + 20 &s=2 $
$latex \Rightarrow (y+1)^3 = (x+1)^3 + 2x^2 - 22x + 20 &s=2 $
We have two cases :
Case 1: $latex 2x^2 - 22x + 20 \le 0 &s=2 $ implying $latex y \le x &s=2 $
$latex 2x^2 - 22x + 20 \le 0 &s=2 $
$latex \Rightarrow x^2 - 11x + 10 \le 0 &s=2 $
$latex \Rightarrow (x-1)(x-10) \le 0 &s=2 $
$latex \Rightarrow 1 \le x \le 10 &s=2 $
This is an easy check (plug in 1, 2, ... , 10).
We find $latex x = 1, x = 10; &s=2 $ works.
Case 2: $latex 2x^2 - 22x + 20 \ge 0 &s=2 $ implying $latex y \ge x &s=2 $
Then y is at least x+1.
$latex (x+1)^3 + 2x^2 - 22x + 20 = (y+1)^3 \ge (x+1+1)^3 = (x+2)^3 &s=2 $
$latex \Rightarrow (x+1)^3 + 2x^2 - 22x + 20 \ge (x+2)^3 &s=2 $
$latex \Rightarrow 2x^2 - 22x + 20 \ge (x+2)^3 - (x+1)^3 &s=2 $
$latex \Rightarrow 2x^2 - 22x + 20 \ge x^3 + 6x^2 + 12x + 1 - x^3 - 3x^2 - 3x - 1 &s=2 $
$latex \Rightarrow 2x^2 - 22x + 20 \ge 3x^2 + 9x -7 &s=2 $
$latex \Rightarrow 0 \ge x^2 + 31x -27 &s=2 $
But this implies $latex \displaystyle{\Rightarrow \frac{-31 - \sqrt{1013}}{2} \le x \le \frac{-31 + \sqrt{1013}}{2}} &s=2 $
But x is a positive integer. Hence this is not possible.
