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.

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