Prove that if a, b, c are integers and a√2 + b√3 + c = 0, then a = b = c = 0.
√3 is irrational. For if not, let √3 = r/s with r and s having no common factor. But now 3s2 = r2, so 3 divides r2 and hence r. Hence 3 divides s2 and hence s. So r and s have a common factor 3. Contradiction. Similarly √2 is irrational.
If a√2 + b√3 + c = 0, then 2a2 - 3b2 - c2 = 2bc √3. But √3 is irrational, so b or c = 0. If b = 0, then a √2 + c = 0. But √2 is irrational, so a = c = 0 also.
Hence c = 0. So a √2 + b √3 = 0. We may take a and b to be relatively prime (divide out any common factor). Squaring: 2a2 = 3b2.
15th Putnam 1955
© John Scholes
26 Nov 1999