Find all real solutions (x, y) to the simultaneous equations:
1/x - 1/(2y) = 2 y4 - 2 x4
1/x + 1/(2y) = (3 x2 + y2)(x2 + 3 y2).
Answer: There is just one solution: x = (31/5 + 1)/2, y = (31/5 - 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 x4, the equations become: 1/x5 - 1/(2 t x5) = 2t4 - 2, 1/x5 + 1/(2 t x5) = 3t4 + 10t2 + 3. It is now easy to eliminate x5 to get a quintic in t: 2t5 - 5t4 + 20t3 - 10t2 + 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(t2 - t + 1)2 + 30t2), 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 = x4 + 10x2y2 + 5y4, 1/y = 5x4 + 10x2y2 + y4. The 1, 5, 10 pattern should look familiar at this point. Multiply across by the x, y to get: 2 = x5 + 10x3y2 + 5xy4, 1 = 5x4y + 10x2y3 + y5. 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 = 31/5.
62nd Putnam 2001
© John Scholes
16 Dec 2001