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
:
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
.
So, and all intervals between
and
will contain at least one perfect square.
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