### 38th Putnam 1977

**Problem A2**

Find all real solutions (a, b, c, d) to the equations a + b + c = d, 1/a + 1/b + 1/c = 1/d.

**Solution**

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 x^{2} - (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

jscholes@kalva.demon.co.uk

30 Nov 1999