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A set of 3 equations (Posted on 2016-12-31) Difficulty: 3 of 5
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 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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