40th Putnam 1979

Problem A5

Show that we can find two distinct real roots α, b of x3 - 10x2 + 29x - 25 such that we can find infinitely many positive integers n which can be written as n = [rα] = [sβ] for some integers r, s.



The polynomial has value -25, -5, 1, -1, -5, -5, 5 at x = 0, 1, 2, 3, 4, 5, 6. So one root lies between 1 and 2, another between 2 and 3, and another between 5 and 6. In particular, all the roots are greater than 1, so if α is a root then all the values [α], [2α], [3α], [4α] ... are different. Hence just [n/α] of them lie in the range 1 to n. Similarly, if the other roots are β, γ, then [n/β] values [mβ] lie in the range 1 to n, and [n/γ] values [mγ]. So at least [n/α] + [n/β] + [n/γ] - n integers in the range 1 to n must have at least two of the representations [mα], [m'β], [m''γ]. But 1/α + 1/β + 1/γ > 1/2 + 1/3 + 1/6 = 1. So as n tends to infinity, the number of integers <= n with at least two representations tends to infinity. In other words, infinitely many positive integers have at least two representations. But there are only finitely many possibilities, so we must be able to find two roots α, β such that infinitely many integers have the representations [mα] and [m'β].

Interestingly, this argument does not tell us which two roots!



40th Putnam 1979

© John Scholes
4 Dec 1999