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A set of 3 equations (Posted on 2016-12-31)

Solve in integers:

x^2 = yz + 1
y^2 = zx + 2
z^2 = xy + 4
P&p solutions only!

No Solution Yet Submitted by Ady TZIDON

Rating: 3.5000 (2 votes)

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Solution

Comment 1 of 1

x² = yz + 1 (1)     y² = zx + 2 (2)     z² = xy + 4 (3)

(2) – (1) gives:             y² – x² = z(x – y) + 1

which simplifies to          (x + y + z)(y – x) = 1     (4)

Similarly, (3) – (2)         (x + y + z)(z – y) = 2     (5)

For integer solutions, (4) & (5) give two possibilities:

Either    x + y + z = 1 and y – x = 1 and z – y = 2,

            from which:       (x, y, z) = (-1, 0, 2)

or         x + y + z = -1 and y – x = -1 and z – y = -2,

            from which:       (x, y, z) = (1, 0, -2)

Sorry, no pencils or paper used…

Posted by Harry on 2016-12-31 09:34:03

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