Вопрос от Анонимного юзера 24 октября 2025 13:56

Question 1

X2+10=y2-6

Answer

Interpret as x^2 + 10 = y^2 - 6.

Rearrange: y^2 - x^2 = 16 (y - x)(y + x) = 16

Let a = y - x and b = y + x. Then ab = 16 and a, b have the same parity. The relevant factor pairs of 16 with same parity are: (2, 8), (4, 4), (8, 2), and their negatives: (-2, -8), (-4, -4), (-8, -2).

Solve x = (b - a)/2, y = (a + b)/2 for each pair:

(a, b) = (2, 8): x = 3, y = 5 → (x, y) = (3, 5)
(4, 4): x = 0, y = 4 → (0, 4)
(8, 2): x = -3, y = 5 → (-3, 5)
(-2, -8): x = -3, y = -5 → (-3, -5)
(-4, -4): x = 0, y = -4 → (0, -4)
(-8, -2): x = 3, y = -5 → (3, -5)

Thus the integer solutions are: (0, 4), (0, -4), (3, 5), (-3, 5), (3, -5), (-3, -5).

If you only want real solutions, then y = ± sqrt(x^2 + 16).

Question 2

X2+10=y2-6

Answer

Interpret as x^2 + 10 = y^2 - 6. Rearrange: y^2 - x^2 = 16 (y - x)(y + x) = 16 Let a = y - x and b = y + x. Then ab = 16 and a, b have the same parity. The relevant factor pairs of 16 with same parity are: (2, 8), (4, 4), (8, 2), and their negatives: (-2, -8), (-4, -4), (-8, -2). Solve x = (b - a)/2, y = (a + b)/2 for each pair: - (a, b) = (2, 8): x = 3, y = 5 → (x, y) = (3, 5) - (4, 4): x = 0, y = 4 → (0, 4) - (8, 2): x = -3, y = 5 → (-3, 5) - (-2, -8): x = -3, y = -5 → (-3, -5) - (-4, -4): x = 0, y = -4 → (0, -4) - (-8, -2): x = 3, y = -5 → (3, -5) Thus the integer solutions are: (0, 4), (0, -4), (3, 5), (-3, 5), (3, -5), (-3, -5). If you only want real solutions, then y = ± sqrt(x^2 + 16).

X2+10=y2-6

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