Solve:
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 multivariable 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 20170222 08:27:17 