1994 USAMO Problems/Problem 1
Let , be positive integers, no two consecutive, and let , for . Prove that, for each positive integer , the interval , contains at least one perfect square.
Solution
We want to show that the distance between and is greater than the distance between and the next perfect square following .
Given , where no are consecutive, we can put a lower bound on . This occurs when all :
\begin{align*} s_n&=(k_{n,min})+(k_{n,min}-2)+(k_{n,min}-4)+\dots+(k_{n,min}-2n+2)\\ &=nk_{n,min}-\sum_{i=1}^{n-1}2i\\ &=nk_{n,min}-2\sum_{i=1}^{n}i+2n\\ &=nk_{n,min}-n(n+1)+2n\\ &=nk_{n,min}-n^2+n \end{align*} (Error compiling LaTeX. Unknown error_msg)
Rearranging, . So, , and the distance between and is .
Also, let be the distance between and the next perfect square following . Let's look at the function for all positive integers .
When is a perfect square, it is easy to see that . Proof: Choose . .
When is not a perfect square, . Proof: Choose with . .
So, for all and for all .
Now, it suffices to show that for all .
\begin{align*} k_{n+1}-d(s_n)&\geq \frac{s_n}{n}+n+1-2\sqrt{s_n}-1\\ &=\frac{1}{n}(s_n+n^2-2n\sqrt{s_n})\\ &=\frac{s_n^2+n^4+2n^2s_n-4n^2s_n}{n(s_n+n^2+2n\sqrt{s_n})}\\ &=\frac{(s_n-n^2)^2}{n(s_n+n^2+2n\sqrt{s_n})}\\ &\geq 0 \end{align*} (Error compiling LaTeX. Unknown error_msg)
So, and all intervals between and will contain at least one perfect square.
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol941.html
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