1994 USAMO Problems/Problem 1

Solution

Induct on $n$ . When $n = 1$ , we are to show that the interval $\, [s_n, s_{n + 1}) \,$ contains at least one perfect square. This interval is equivalent to $\, [k_0, k_0 + k_1) \,$ when $n = 1$ . Now for some $a , a^2 \le k_0^2 < (a+1)^2$ .Then it suffices to show that the minimal "distance spanned" by the interval $\, [k_0, k_0 + k_1) \,$ is greater than or equal to the maximum distance from $k_0$ to the nearest perfect square. Since the smallest element that can follow $k_0$ is $k_0 + 2$ , we have to show the below. Note that we ignore the trivial case where $k_0 = a^2$ , which should be mentioned.

  
  
  \{1, 2, \ldots, 2\a}$$ (Error compiling LaTeX. Unknown error_msg)We now prove by contradiction. Assume that

  $2a > a^2 + q + 1
  a^2 - 2a + (q + 1) < 0
  No real roots (Discriminant < 0)
  =><=

Now assume that this property should be true for$ (Error compiling LaTeX. Unknown error_msg)\, [s_{n - 1}, s_{n}) \, $. Now to show for$ \, [s_n, s_{n + 1}) \,$, 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.

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1994 USAMO Problems/Problem 1

Solution