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 Integer triplets (Posted on 2017-02-21)
 Solve:

x^2 + 2yz+ 5x = 2
y^2 + 2zx+ 5y = 2
z^2 + 2xy+ 5z = 2

 See The Solution Submitted by Ady TZIDON No Rating

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 solution | Comment 1 of 4
Symmetry suggests checking if x=y=z is a solution.  That would mean the quadratic x^2 + 2X^2 + 5x = 2 has an integer solution and indeed, x = -2 works.

For the rest, multiply the first equation by x and the second by y and subtract to get rid of the nasty multi-variable terms.

x^3 - y^3 + 5(x^2 - y^2) = 2(x - y)

Division by (x - y) is allowed and gives a quadratic in x with discriminant -3y^2 - 10y + 33 which is positive for -5 <= y <= 2 and square for y = 2, -3, -4 but only (2,2,-3) is a solution.

Solutions are (-2,-2,-2) and (2,2,-3) and permutations.

 Posted by xdog on 2017-02-22 08:27:17

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