47th Putnam 1986

Problem B4

For real r > 0, define m(r) =min{ |r - √(m2+2n2)| for m, n integers}. Prove or disprove: (1) limr→∞ m(r) exists; and (2) is zero.



Answer: true, true.

Fairly easy.

Two heuristic arguments. First, if we think of lattice points at (m, n/√2), it seems likely that large circles will pass increasingly close to lattice points. So the results look true.

Second, let xm,n = √(m2+2n2), and let us confine our attention to the values xm,n lying between m and m + 1. There are about √m of them in a distance 1 and roughly equally spaced, so any real number between m and m+1 should lie within about 1/√m of one of them.

In making this rigorous, the important thing is to keep it simple. We do not need a good approximation, just an adequate one. Observe first that √(1 + x) < 1 + x/2, so √(a + x) - √a < x/(2 √a). So given any r > 0, take r ∈ (m, m+1], and the largest n st m2 + 2n2 < r2, so that r ∈ ( √(m2 + 2n2), √(m2 + 2(n+1)2) ]. The distance of r from the nearest lattice point is certainly less than the width of this interval, d. Also 2n2 < 2m + 1 < 8m, so n < 2√m. So, using our earlier observation, d < (2n + 1)/√(m2 + 2n2) < (2n + 1)/m < (4√m + 1)/m which tends to zero as r → ∞.



47th Putnam 1986

© John Scholes
3 Oct 1999