Can we find n such that μ(n) = μ(n + 1) = ... = μ(n + 1000000) = 0? [The Möbius function μ(r) = 0 iff r has a square factor > 1.]
Let pn be the nth prime. We show how to find an such that p12 divides an, p22 divides an + 1, p32 divides an + 2, ... , pn2 divides an + n - 1.
Evidently we can take a1 = 4. We now use induction. an + k p12 ... pn2 has the same property as an. So we need to pick k such that an + k p12 ... pn2 = 0 (mod pn+12). But that is always possible (by Euclid's algorithm, for example) since pn+12 is relatively prime to p12 ... pn2.
15th Putnam 1955
© John Scholes
26 Nov 1999