Find all real solutions (a, b, c, d) to the equations a + b + c = d, 1/a + 1/b + 1/c = 1/d.
Answer: a, b arbitary; c = - a, d = b.
Take a, b arbitary. We then have: -c + d = a + b; cd = -ab. So -c and d are the roots of the quadratic x2 - (a + b)x + ab = 0. Solving, the roots are a, b. So either c = -a, d = b, or c = -b, d = a.
38th Putnam 1977
© John Scholes
30 Nov 1999