2008 USAMO Problems/Problem 1
Problem
(Titu Andreescu)
Prove that for each positive integer , there are pairwise relatively prime integers
, all strictly greater than 1, such that
is the product of two consecutive integers.
Solutions
Solution 1
We will prove the problem for each nonnegative integer . We wish to show that
for some integer
. We induct on
. For our base case,
, we may let
be positive integer.
For the inductive step, suppose that are pairwise relatively prime integers such that
for some integer
. Let
. Evidently,
. Also,
Since
is odd and relatively prime to
, it follows that
and
are relatively prime, so
is relatively prime to each of
. Finally,
This completes the induction.
Solution 2
Lemma. If is prime such that
, there exists a residue
such that
.
Proof. Let be a multiplicative generator of the nonzero integers mod p. Set
. Then
, but
, so
.
By Dirichlet's Theorem, there are infinitely many primes congruent to 1 (mod 3). Let be
such primes, and let
be respective residues as described in the lemma. By the Chinese Remainder Theorem, there is a positive integer
that satisfies the relation
for each integer
. Then
Now, for
, take
to be the greatest power of
that divides
, and let
. Since all the
are pairwise relatively prime and are greater than 1, we are done.
Solution 3
Firstly, we see that there are numbers
. Since
, there are at least 2 values of
.
Define a relatively prime partition to be a set of relatively prime numbers such that their product is equal to some natural number.
Define to be the greatest possible cardinality of a relatively prime partition for that number.
Lemma 1: All cardinalities of the relatively prime partitions of a number up to can be attained.
Proof:
satisfies the properties. For any
which satisfies the properties, we can take any of the 2
numbers and multiply them together. Because they are both relatively prime to all the other numbers, their product is relatively prime to all the other numbers as well, and that results in
numbers which satisfy the conditions, unless
, because there is only one number left. Therefore, all numbers of numbers of relatively prime factors from
to
are attainable, if
is attainable as well.
End Lemma
Lemma 2: can have arbitrarily many prime factors for some value of
.
Proof:
Let . Let
. Then
.
After factoring, we get
.
, because
is strictly positive.
Let their GCD=g.
, and so
, but
which is strictly odd, so g=1.
therefore must contain prime factors not in
. Upon repeating this an arbitrary number of times, we get a number of the form
which has arbitrarily many distinct prime factors.
End Lemma
Then would not have
elements in the relatively prime partition for some value of n, and any value of a.
It is possible to choose a value of
such that
has arbitrarily many unique prime factors, by our second Lemma, and so it is possible for
to be arbitrarily high. By our first Lemma, all numbers up to P are possible values for the cardinality of some relatively prime partition, and so there always exists some number that has an arbitrary number of elements in a relatively prime partition. Because
,
is the product of 2 consecutive integers, we see that the given statement is true because if
, then
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
2008 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |