Putnam 1994 Problem B1 For a positive integer n, let f(n) be the number of perfect squares d such that |n - d| ≤ 250. Find all n such that f(n) = 15. [The perfect squares are 0, 1, 4, 9, 16, ... .] Solution
: 315 - 325 and 332 - 350. Straightforward. We may readily compile the following table: 0 -250 ≥ 16 0 2 - 15 2 251-299 ≥ 16 7 2 - 22 2 300-314 16 8 2 - 23 2 315-325 15 9 2 - 23 2 326-331 16 9 2 - 24 2 332-350 15 10 2 - 24 2 351-371 14 11 2 - 24 2 25 2 - 11 2 = 504 > 500 and hence the gap (n+15) 2 - (n+1) 2 > 500 for all n > 10. So no numbers n > 11 2 + 250 = 371 have f(n) > 14. Putnam 1994 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998