Difference between revisions of "1994 USAMO Problems/Problem 1"

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Induct on <math>n</math>. When <math>n = 1</math>, we are to show that the interval <math>\, [s_n, s_{n + 1}) \,</math> contains at least one perfect square. This interval is equivalent to <math>\, [k_0, k_0 + k_1) \,</math> when <math>n = 1</math>. Now for some <math>a , a^2 \le k_0^2 < (a+1)^2</math>.Then it suffices to show that the minimal "distance spanned" by the interval <math>\, [k_0, k_0 + k_1) \,</math> is greater than or equal to the maximum distance from <math>k_0</math> to the nearest perfect square. Since the smallest element that can follow <math>k_0</math> is <math>k_0 + 2</math>, we have to show the below. Note that we ignore the trivial case where <math>k_0 = a^2</math>, which should be mentioned.
  <math>2a + 1 \le k_0 + 2</math>
  <math>2a \le k_0 + 1</math>
  <math>2a \le k_0 + (q + 1), where q is a member of </math>\{1, 2, \ldots, 2\a}<math>
  </math>We now prove by contradiction. Assume that
  <math>2a > a^2 + q + 1
  a^2 - 2a + (q + 1) < 0
  No real roots (Discriminant < 0)
  =><= </math>
<math>Now assume that this property should be true for </math>\, [s_{n - 1}, s_{n}) \,<math>. Now to show for </math>\, [s_n, s_{n + 1}) \,<math>, we note that an interval is defined solely by the sum of the elements in the respective sets and '''not''' on the length of the set. Thus any s_{n + 1} can be turned into an s_n by removing k_{n + 1} and adding it to k_n. The new set is now in a form to which we can apply our assumption. The same goes for s_n and s_{n - 1}. Thus we have finished the inductive argument.
Therefore etc.</math>

Revision as of 06:26, 11 July 2009


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