Find all real solutions (x, y) to the simultaneous equations:

1/x - 1/(2y) = 2 y^{4} - 2 x^{4}

1/x + 1/(2y) = (3 x^{2} + y^{2})(x^{2} + 3 y^{2}).

**Solution**

Answer: There is just one solution: x = (3^{1/5} + 1)/2, y = (3^{1/5} - 1)/2.

We need to find some way of eliminating one of the variables. The rhs of both equations is homogeneous, so this suggests using the variable t = y/x. Dividing through by x^{4}, the equations become: 1/x^{5} - 1/(2 t x^{5}) = 2t^{4} - 2, 1/x^{5} + 1/(2 t x^{5}) = 3t^{4} + 10t^{2} + 3. It is now easy to eliminate x^{5} to get a quintic in t: 2t^{5} - 5t^{4} + 20t^{3} - 10t^{2} + 10t - 1 = 0.

However, it is not clear that this leads anywhere, if we differentiate we find that the derivative is the sum of two squares (10(t^{2} - t + 1)^{2} + 30t^{2}), so the quintic is monotonic increasing and has just one real root. But a little trial makes clear that the root is not anything particularly nice (it lies between 0 and 1, but is not 1/2).

The trick is to play around until you recognize the binomial coefficients for n = 5. If you multiply out the second equation and then add and subtract the two equations, you get: 2/x = x^{4} + 10x^{2}y^{2} + 5y^{4}, 1/y = 5x^{4} + 10x^{2}y^{2} + y^{4}. The 1, 5, 10 pattern should look familiar at this point. Multiply across by the x, y to get: 2 = x^{5} + 10x^{3}y^{2} + 5xy^{4}, 1 = 5x^{4}y + 10x^{2}y^{3} + y^{5}. Adding and subtracting gives: 3 = (x + y)^{5}, 1 = (x - y)^{5}.

We want real solutions, so x + y = k, x - y = 1, where k = 3^{1/5}.

© John Scholes

jscholes@kalva.demon.co.uk

16 Dec 2001