Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Bulgaria

Find all pairs $(x, y)$ of real numbers for which
$4y^4 + x^4 + 12y^3 + 5x^2(y^2 + 1) + y^2 + 4 = 12y.$

Solutions — 2

Solution 1

We have the inequalities $x^4 \geq 0$ , $5x^2(y^2+1) \geq 0$ and $4y^4 + 12y^3 + y^2 - 12y + 4 = (2y-1)^2(y+2)^2 \geq 0$ . The sum of the left sides is 0 if and only if each of them is equal to 0. The first two lead to $x = 0$ , and the third to $y = -2$ or $y = \frac{1}{2}$ . $\square$

Solution 2

We have the inequalities $x^4 \geq 0$ , $5x^2(y^2 + 1) \geq 0$ and $4y^4 + 12y^3 + y^2 - 12y + 4 = (2y - 1)^2(y + 2)^2 \geq 0$ . The sum of the left sides is $0$ if and only if each of them is equal to $0$ . The first two lead to $x = 0$ , and the third to $y = -2$ or $y = \frac{1}{2}$ .

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