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

**Solution**

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 x_{m,n} = √(m^{2}+2n^{2}), and let us confine our attention to the values x_{m,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 m^{2} + 2n^{2} < r^{2}, so that r ∈ ( √(m^{2} + 2n^{2}), √(m^{2} + 2(n+1)^{2}) ]. The distance of r from the nearest lattice point is certainly less than the width of this interval, d. Also 2n^{2} < 2m + 1 < 8m, so n < 2√m. So, using our earlier observation, d < (2n + 1)/√(m^{2} + 2n^{2}) < (2n + 1)/m < (4√m + 1)/m which tends to zero as r → ∞.

© John Scholes

jscholes@kalva.demon.co.uk

3 Oct 1999