1972 IMO Problems/Problem 3
Contents
[hide]Problem
Let and
be arbitrary non-negative integers. Prove that
is an integer. (
.)
Solution 1
Denote the given expression as . We intend to show that
is integral for all
. To start, we would like to find a recurrence relation for
. First, let's look at
:
Second, let's look at
:
Combining,
Therefore, we have found the recurrence relation
Note that is just
, which is an integer for all
. Then
so
is an integer, and therefore
must be an integer, etc.
By induction, is an integer for all
.
Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html
Solution 2
Let p be a prime, and n be an integer. Let be the largest positive integer
such that
WTS: For all primes ,
We know
Lemma 2.1: Let be real numbers. Then
Proof of Lemma 2.1: Let and
On the other hand,
It is trivial that (Triangle Inequality)
Apply Lemma 2.1 to the problem: and we are pretty much done.
Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.
Remark about, and correction to Solution 2
The last step seems mistaken. For example [1.5]+[1.5]<[1.5+1.5]. Triangle inequality is about absolute value |x| not floor function [x]. Following is a rectified solution.
Lemma: [a+b]<=[a]+[b]+1, a,b>=0.
Proof: a<[a]+1, b<[b]+1, so a+b<[a]+[b]+2, [a+b]<=a+b<[a]+[b]+2, so [a+b]<=[a]+[b]+1.
Let (x,2x,…,nx)=n, p is any prime, V2(9!)=V2(1•2•3•4•5•6•7•8•9) =(2,4,6,8)+(4,8)+(8)=[9/2]+[9/4]+[9/8]=7.
Generalize it and we have, Vp(m!)=Vp(1•2•….•m)=[m/p]+[m/p^2]+…+[m/p^k], k=1~∞, or,
Vp(m!)=∑([m/p^k],k=1~∞).
Obviously when p^k>m, k>=1, [m/p^k]=0.
Note the target statement is equivalent to:
for any prime p, Vp((2m)!(2n)!)>=Vp(m!n!(m+n)!), or,
Vp((2m)!)+Vp((2n)!)>=Vp(m!)+Vp(n!)+Vp(m+n)! .
Let m/p^k=a(k), n/p^k=b(k), k=1~∞. Combine (1), we note that (2) is equivalent to:
for any k in (1), [2a(k)]+[2b(k)]>=[a(k)]+[b(k)]+[a(k)+b(k)] .
For convenience of viewing, let a(k)=a, b(k)=b, a,b>=0, (3) is then rewritten as:
[2a]+[2b]>=[a]+[b]+[a+b] .
We prove it under different scenarios:
-If a<[a]+1/2, b<[b]+1/2, Then 2[a]<=[2a]<=2a<2[a]+1, [2a]=2[a], likewise, [2b]=2[b], so (4) is equivalent to [a]+[b]>=[a+b]. Per current condition, a+b<[a]+[b]+1, so [a+b]<[[a]+[b]+1]=[a]+[b]+1, also obviously [a+b]>=[a]+[b], so [a+b]=[a]+[b], proved;
-If a<[a]+1/2, b>=[b]+1/2, then [2a]=2[a], [2b]=2[b]+1, so (4) is equivalent to [a]+[b]+1>=[a+b], as per lemma, proved;
-If a>=[a]+1/2, b>=[b]+1/2, so [2a]=2[a]+1, [2b]=2[b]+1, so (4) is equivalent to [a]+[b]+2>=[a+b], as per lemma, proved;
Therefore (4) is proved, so is (2), as every step from (2) to (4) is reversible. Therefore (2m)!(2n)!/(m!n!(m+n)!) is an integer. Proof completed.
Remarks (added by pf02, April 2025)
1. The first solution has a few errors in the computation, but they cancel out, and the proof is essentially correct. Below I will give the corrections, to make the solution flawless.
2. As pointed out by a contributor, the second solution is incorrect. The correction given is good, but very sloppily written and hard to read (even after I edited the spacing, to make it legible). Below, I will repeat the argument in a simplified form, to make it easier to follow.
3. I will give a third solution below, inspired by the papers mentioned in the next paragraph.
4. The problem is a classical result, attributed to Eugene
Catalan, who stated it in a paper published in 1874. The
numbers are
sometimes called the super-Catalan numbers. (They are
mentioned at the end of
https://en.wikipedia.org/wiki/Catalan_number).
There are a few papers published about them. See for example
the paper "The Super Catalan Numbers for
and Some Integer Factorial Ratios" by Xin Chen & Jane Wang
(https://www-users.cse.umn.edu/~reiner/REU/ChenWang2012.pdf),
and "Super Ballot Numbers" by Ira M. Gessel (especially section 6)
(https://www.sciencedirect.com/science/article/pii/0747717192900342/pdf).
Corrected computations in Solution 1
We have
and
The rest of the proof is correct.
Simplified version of the correction to Solution 2
To prove that is an
integer, we have to prove that for any prime number
, the
exponent of the largest power of
in
is larger
than the exponent of the largest power of
in
.
(In fact, we have to consider only primes which satisfy
).
The largest power of in
! is given by
.
(The sum is a finite sum, since we have to consider only powers
This is easy to see, but it
is a reasonably well known fact (see Legendre's Formula or
https://en.wikipedia.org/wiki/Legendre's_formula.) Note that
.
Thus, it is enough to prove that
.
Let be the integer division of
by
(so
), and let
. In other words,
.
Note that
. Similarly, obtain
from the division
of
by
.
Note that
or
depending on whether
or
and similarly
for
and
. Also
or
depending on whether
or
.
We have the following cases:
1. If and
then we have
equality in (1).
2. If , and
then after cancelling terms, the left hand side in (1) becomes 1, and
the right hand side becomes 0.
3. If , and
then we have equality in (1).
4, 5. The two cases when
are similar to cases 2 and 3.
6. If , then after cancelling
terms, the left hand side in (1) becomes 2, and the right hand side
becomes 1.
The above shows that (1) is always true, which proves the problem.
Solution 3
This solution uses the fact that
are the coefficients
of the powers of
in the expansion of
(see Binomial Theorem
or https://en.wikipedia.org/wiki/Binomial_coefficient). Of course these
coefficients, hence these expressions in
are integers.
Without loss of generality, we can assume . We will use the identity
.
This identity is attributed to K. von Szily (1894). Note that we could
take the sum over all integers, using the convention that
when
or
.
The fact that is integer follows immediately from this identity
because the right hand side is integer.
Thus, we just need to prove this identity.
Start with . Expand
both sides and look at the coefficient of
in both expansions.
Equate the two expressions giving the coefficient of
in the two
sides. We get
.
Write out the binomial coefficients in terms of factorials, and multiply
both sides by . We get
Simplifying and rearranging factors, we get
This finishes the problem.
See Also
1972 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.