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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.