Math Olympiad | A Nice Exponential Problem | 90% Failed to solve! 

The constant term in this quartic equation is (-2)² - 2 = 2. This being the case, if there are rational roots, then they are x = ±1 or ±2. By observation, x = -1 and x = 2 are in fact roots. We can expand out the quartic and then divide it by (x + 1)(x - 2) = x² - x - 2 to find the remaining quadratic factor, which can easily be solved.

(x² - 2)² = x + 2 x⁴ - 4x² + 4 = x + 2 x⁴ - 4x² + 4 - x - 2 = 0 x⁴ - 4x² - x + 2 = 0 ← it would be interesting to have 2 squares on the left side (because power 4 and power 2) Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce 2 squares on the left side Let's tinker a bit with x⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ² x⁴ - 4x² - x + 2 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ² (x² + λ)² - 2λx² - λ² - 4x² - x + 2 = 0 (x² + λ)² - [2λx² + λ² + 4x² + x - 2] = 0 → let's try to get [...] as a square (x² + λ)² - [x².(2λ + 4) + x + (λ² - 2)] = 0 → a square into […] means that Δ = 0 → let's calculate Δ Δ = (1)² - 4.[(2λ + 4).(λ² - 2)] → then, Δ = 0 (1)² - 4.[(2λ + 4).(λ² - 2)] = 0 4.[(2λ + 4).(λ² - 2)] = 1 8.[(λ + 2).(λ² - 2)] = 1 (λ + 2).(λ² - 2) = 1/8 λ³ - 2λ + 2λ² - 4 - (1/8) = 0 λ³ + 2λ² - 2λ - (33/8) = 0 λ = - 3/2 Restart (x² + λ)² - [x².(2λ + 4) + x + (λ² - 2)] = 0 → when λ = - 3/2, a square will appear [x² - (3/2)]² - [x².(2.{- 3/2} + 4) + x + ({- 3/2}² - 2)] = 0 [x² - (3/2)]² - [x².(- 3 + 4) + x + ({9/4} - 2)] = 0 [x² - (3/2)]² - [x² + x + (1/4)] = 0 ← we can see a square [x² - (3/2)]² - [x + (1/2)]² = 0 → recall: a² - b² = (a + b).(a - b) { [x² - (3/2)] + [x + (1/2)] }.{ [x² - (3/2)] - [x + (1/2)] } = 0 [x² - (3/2) + x + (1/2)].[x² - (3/2) - x - (1/2)] = 0 [x² + x - (2/2)].[x² - x - (4/2)] = 0 (x² + x - 1).(x² - x - 2) = 0 First case: (x² + x - 1) = 0 x² + x - 1 = 0 Δ = (1)² - (4 * - 1) = 5 x = (- 1 ± √5)/2 → x = (- 1 + √5)/2 → x = (- 1 - √5)/2 Second case: (x² - x - 2) = 0 x² - x - 2 = 0 Δ = (- 1)² - (4 * - 2) = 9 x = (1 ± 3)/2 → x = 2 → x = - 1

(x^4 ➖ 4)={x^0+x^0 ➖ }=x^1 (x ➖ 1x+1).{x+x ➖ }+{2+2 ➖}={x^2+4}=4x^2 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).