IMO shortlist 1996/N8 solution

IMO shortlist 1996

Problem N8

For each odd prime p, find positive integers m, n such that m ≤ n and (2p)^1/2 - m^1/2 - n^1/2 is non-negative and as small as possible.

Solution

Answer: (p-1)/2 and (p+1)/2.

We show first that we cannot get zero.

Suppose (2p)^1/2 - m^1/2 - n^1/2 = 0. Then 2p = m + n - 2(mn)^1/2, so 4p² + (m+n)² - 4p(m+n) = 4mn, so 4p² + (m-n)² -4p(m+n) = 0. Hence p divides (m-n)². But p is prime, so p must divide (m-n) and hence p² divides (m-n)². Hence p divides (m+n). Hence p divides 2m and 2n. But p is odd, so p divides m and n. If m ≥ 2p, then n ≤ 0, but n is positive, so m < 2p. But m is positive, so m = p. Similarly, n = p. But (2p)^1/2 - p^1/2 - p^1/2 is not zero. Contradiction. So we cannot get zero.

Put D = (2p)^1/2 - m^1/2 - n^1/2. Multiplying and dividing by ( (2p)^1/2 + m^1/2 + n^1/2) we get D = (2p - m - n - 2(mn)^1/2)/( (2p)^1/2 + m^1/2 + n^1/2). Multiplying and dividing by (2p - m - n + 2(mn)^1/2) we get D = ( (2p - m - n)² - 4mn)/( ( (2p)^1/2 + m^1/2 + n^1/2)(2p - m - n + 2(mn)^1/2) ).

Suppose the numerator is 1. Then (2p - m - n)² = 4mn + 1. So 2p - m - n = 2r + 1, where 4mn + 1 = (2r + 1)². Hence mn = r(r+1). Let d be the greatest common divisor of m and r, so that m = Md, r = Rd, with M and R coprime. Then M must divide r+1. Put r+1 = Md'. Then Mdn = RdMd', so n = Rd'. Hence 2p = m + n + 2r + 1 = Md + Rd' + Md' + Rd = (M+R)(d+d'). But p is an odd prime, so either d = d' = 1 or M = R = 1. If d = d' = 1, then m = M = r+1 = R+1 = n+1. But m <= n. So we must have M = R = 1. So m = d, r = d, d' = d+1, n = d+1, p = 2d+1. Hence D = 1/( ( (4d+2)^1/2 + d^1/2 + (d+1)^1/2)(2d+1 + 2(d²+d)^1/2) ) < 1/( (4d)^1/2 + d^1/2 + d^1/2)(2d + 2(d²)^1/2) ) = 1/( 4d^1/2 4d) = 1/(16d^3/2).

Now suppose the numerator is greater than 1. Then it must be at least 2. Looking at the denominator, we have (2p)^1/2 - m^1/2 - n^1/2 > 0 and hence (2p)^1/2 + m^1/2 + n^1/2 < 2 (2p)^1/2. Also (2p - m - n + 2(mn)^1/2) = (2p - (m^1/2 - n^1/2)²) < 2p. So D > 2/( 2 (2p)^1/2 2p) = 1/(2p)^3/2. So if we put p = 2d+1, then 2p = 4d+2 <= 6d < 16^2/3d, so D > 1/(16d^3/2).

Thus we have D > 1/16d^3/2 for all m, n except m = (p-1)/2, n = (p+1)/2 when we have D < 1/16d^3/2.

37th IMO shortlist 1996