x^2 + 2yz+ 5x = 2
y^2 + 2zx+ 5y = 2
z^2 + 2xy+ 5z = 2
Pen and paper only, please.
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