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.
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