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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.

Chatuspathi:

December 31, 2015

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