IMO 1984 SL 3

Origin: USS

Problem

Find all positive integers n such that n = d2 6 + d2 7 −1, where 1 = d1 < d2 < \cdot \cdot \cdot < dk = n are all positive divisors of the number n.

Solution

From the equality n = d2 6 + d2 7 −1 we see that d6 and d7 are relatively prime and d7 | d2 6 −1 = (d6 −1)(d6 + 1), d6 | d2 7 −1 = (d7 −1)(d7 + 1). Suppose that d6 = ab, d7 = cd with 1 < a < b, 1 < c < d. Then n has 7 divisors smaller than d7, namely 1, a, b, c, d, ab, ac, which is impossible. Hence, one of the two numbers d6 and d7 is either a prime p or the square of a prime p2, where p is not 2. Let it be di, i \in{6, 7}; then di | (dj−1)(dj+1) implies that dj \equiv\pm1 (mod di), and consequently (d2 i −1)/dj \equiv\pm1 as well. But either dj or (d2 i −1)/dj is less than di, and therefore equals di −1. We thus conclude that d7 = d6 + 1. Setting d6 = x, d7 = x + 1 we obtain that n = x2 + (x + 1)2 −1 = 2x(x + 1) is even. (i) Assume that one of x, x + 1 is a prime p. The other one has at most 6 divisors and hence must be of the form 23, 24, 25, 2q, 2q2, 4q, where q is an odd prime. The numbers 23 and 24 are easily eliminated, while 25 yields the solution x = 31, x + 1 = 32, n = 1984. Also, 2q is eliminated because n = 4pq then has only 4 divisors less than x; 2q2 is eliminated because n = 4pq2 has at least 6 divisors less than x; 4q is also eliminated because n = 8pq has 6 divisors less than x. (ii) Assume that one of x, x+1 is p2. The other one has at most 5 divisors (p excluded), and hence is of the form 23, 24, 2q, where q is an odd prime. The number 23 yields the solution x = 8, x + 1 = 9, n = 144, while 24 is easily eliminated. Also, 2q is eliminated because n = 4p2q has 6 divisors less than x. Thus there are two solutions in total: 144 and 1984.