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Find all natural numbers x > 1 for which x^2 - 3 is divisible by x-1

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Answer: x = { 2 ; 3 }

+) x² - 3 = x² - 1 - 2 = (x - 1)(x + 1) - 2

+) x² - 3 is divisible by x-1 but (x -1)(x + 1) is divisible by x-1

=> 2 must be divisible by x-1

=> x - 1 is divisor of 2

=> (x - 1) ∈ {-1; 1; 2; - 2;}

but x > 1 => x - 1 > 0

⇒ x - 1 = {1; 2; }

+) with x - 1 = 1 => x = 2

x - 1 = 2 => x = 3

Explanation:

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