1986 USAMO Problems/Problem 3
Let's first obtain an algebraic expression for the root mean square of the first integers, which we denote
. By repeatedly using the identity
, we can write
and
We can continue this pattern indefinitely, and thus for any
positive integer
,
Since
, we obtain
Therefore,
Requiring that
be an integer, we find that
where
is an integer. Using the Euclidean algorithm, we see that
, and so
and
share no
factors greater than 1. The equation above thus implies that
and
is each proportional to a perfect square. Since
is
odd, there are only two possible cases:
Case 1: and
, where
and
are integers.
Case 2: and
.
In Case 1, . This means that
for some integers
and
. We proceed by checking whether
is a perfect square for
. (The solution
leads to
, and we are asked to find a value of
greater than 1.) The smallest positive integer
greater than 1 for
which
is a perfect square is
, which results in
.
In Case 2, . We proceed by checking whether
is a perfect square for
. We find that
is not a perfect square for
, and
when
. Thus the smallest positive integers
and
for which
result in a value of
exceeding the value found in Case 1, which was 337.
In summary, the smallest value of greater than 1 for which
is an integer is
.