IMO 2003 SL N4

Origin: ROM | Category: Number Theory

Problem

Let b be an integer greater than 5. For each positive integer n, consider the number xn = 11 . . .1    n−1 22 . . .2    n 5, written in base b. Prove that the following condition holds if and only if b = 10: There exists a positive integer M such that for every integer n greater than M, the number xn is a perfect square.

Solution

Assume that b \geq6 has the required property. Consider the sequence yn = (b −1)xn. From the definition of xn we easily find that yn = b2n + bn+1 + 3b −5. Then ynyn+1 = (b −1)2xnxn+1 is a perfect square for all n > M. Also, straightforward calculation implies  b2n+1 + bn+2 + bn+1 −b3 2 < ynyn+1 <  b2n+1 + bn+2 + bn+1

• b3 2 . Hence for every n > M there is an integer an such that |an| < b3 and ynyn+1 =
  b2n + bn+1 + 3b −5 
  b2n+2 + bn+2 + 3b −5  =  b2n+1 + bn+1(b + 1)
• an 2 . (1) Now considering this equation modulo bn we obtain (3b −5)2 \equiva2 n, so that assuming that n > 3 we get an = \pm(3b −5). If an = 3b −5, then substituting in (1) yields 1 4b2n(b4 −14b3 + 45b2 − 52b + 20) = 0, with the unique positive integer solution b = 10. Also, if an = −3b + 5, we similarly obtain 1 4b2n(b4 −14b3 −3b2 + 28b + 20) − 2bn+1(3b2 −2b −5) = 0 for each n, which is impossible. For b = 10 it is easy to show that xn =
  10n+5 2 for all n. This proves the statement. Second solution. In problems of this type, computing zn = \sqrtxn asymp- totically usually works. From limn\to\infty b2n (b−1)xn = 1 we infer that limn\to\inftybn zn = \sqrt b −1. Further- more, from (bzn + zn+1)(bzn −zn+1) = b2xn −xn+1 = bn+2 + 3b2 −2b −5 we obtain lim n\to\infty(bzn −zn+1) = b \sqrt b −1 . Since the zn’s are integers for all n \geqM, we conclude that bzn −zn+1 = b \sqrt b−1 for all n sufficiently large. Hence b −1 is a perfect square, and moreover b divides 2zn+1 for all large n. It follows that b | 10; hence the only possibility is b = 10.