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Factorials Can't Be Squares

Date: 02/11/2000 at 13:20:25
From: Animesh Datta
Subject: Number Theory

Prove that the factorial of a number (greater than 1) can never be a 
perfect square.

Date: 02/11/2000 at 13:35:07
From: Doctor Wilkinson
Subject: Re: Number Theory

Hi, Animesh,

This follows from Bertrand's Postulate, which states that there is 
always a prime between m and 2m. From this you can conclude that there 
is a prime that divides n! but whose square does not. Since in the 
prime decomposition of a perfect square every exponent must be even, 
this shows that n! cannot be a perfect square.

Here's a sketch of a proof of Bertrand's Postulate:

The proof of Bertand's Postulate uses some simple properties of the 
function theta(x), defined for x >= 0 by

     theta(x) = sum(log p: p is prime and 0 < p <= x)

We show that

     theta(x) < 2x log(2)

(I use log(x) always to mean the natural log of x.)

It is enough to show this when x is an integer.  We're going to prove 
this by induction.

The trick is to look at the binomial coefficient C(2m+1,m), which is


Call this M for short.

Let p be a prime such that m+1 < p <= 2m+1. Then p divides the 
numerator of M but not the denominator, so p divides M. So the product 
of all such primes divides M, and

     sum (log p: m+1 < p <= 2m+1) < log M

or in terms of the function theta(x)

     theta(2m+1) - theta(m+1) < log M.

On the other hand, the binomial expansion of (1 + 1)^(2m+1) has two 
terms equal to M, so

        2M < 2^(2m+1)

         M < 2^2m

     log M < 2m log(2)


     theta(2m+1) - theta(m+1) < 2m log(2)

We're going to use this formula in the induction step of our proof 

     theta(x) < 2x log(2)

For x = 1, we have

     theta(1) = 0 < 2 log(2)

and for x = 2, we have

     theta(2) = log 2 < 4 log(2)

Suppose the inequality is true for x < n. Let us prove it for x = n.

If n is even and > 2, then it is certainly not prime, so

     theta(n) = theta(n-1) < 2(n-1) log(2) < 2n log(2).

If n is odd, let n = 2m + 1.  Then by what we proved above, we have

     theta(2m+1) - theta(m+1) < 2m log(2)

     theta(2m+1) < theta(m+1) + 2m log(2)

                 < 2(m+1) log(2) + 2m log(2)

                 = (3m + 1) log(2)

                 < (4m + 2) log(2)

                 = 2n log(2).

This completes the proof that

     theta(x) < 2x log(2).

Let's catch our breath.

The next thing we're going to do is to look at the highest power of p 
that divides n!, where p is any prime. We call this number j(n, p).

We use the notation [x] for the largest integer <= x.

Every p'th number is a multiple of p, so we get [n/p] factors of p in 
n!. But every p^2'th number is a multiple not just of p but of p^2, 
and [n/p] doesn't count these, so we need to add [n/p^2] for these 
extra factors of p. Similarly every p^3'th number is a multiple of p^3 
which we have not counted yet. So the highest power of p that divides 
n! is the sum of all the


for m >= 1. Of course [n/p^m] = 0 as soon as p^m > n: that is, for 
m > log(n)/log(p).

Now we're going to suppose that Bertrand's Postulate is false, and 
that there is no prime p such that n < p < 2n, for some n.

We're going to look at another binomial coefficient. This one is

     C(2n,n) = (2n!)/(n!)^2

which we'll call N for short.

By our assumption, all the primes that divide N are <= n.  Now using 
the notation above, we have

     N = (2n)!/(n!)^2

         product(p^j(2n, p): p <= 2n)
       = ----------------------------
         product(p^2j(n, p): p <= n)

but there aren't any primes between n and 2n by assumption, so the 
"p <= 2n" in the numerator can be replaced by "p < = n" and we get

     N = product (p^(j(2n, p) - 2j(n, p)): p <= n).

Let's call j(2n, p) - 2j(n, p) k(p) for short. Taking logs on both 
sides, we get

     log N = sum(k(p) log(p): p <= n).

Notice that k(p) is a sum of terms of the form [2x] - 2[x]. 
[2x] - 2[x] is always either 0 or 1. If [2x] is even, [2x] - 2[x] is 
0; otherwise it is 1.

We show first that k(p) = 0 for p > 2n/3.  For in that case,

     2n/3 < p <= n

or   2 <= 2n/p < 3

and [2n/p] = 2, so [2n/p] - 2[n/p] = 0.

     p^2 > (4/9)n^2 > 2n as long as n > 4,

and we can certainly assume n is > 4, since we are assuming there is 
no prime between n and 2n, and 5, for example, is between 4 and 8.

So there are no terms involving higher powers of p.

Next we show that terms with k(p) >= 2 don't contribute very much.

To get such a term we have to have p^2 < 2n or p < sqrt(2n), so the 
number of such terms is at most sqrt(2n).  k(p), on the other hand, is 
a sum of terms [2n/p^m] - 2[n/p^m], which is certainly 0 if p^m > 2n, 
or m > log(2n)/log(p), so k(p) is at most log(2n)/log(p),and 
k(p)log(p) <= log(2n), so

     sum(k(p)log(p) : k(p) >= 2) <= sqrt(2n) log(2n)

taking the maximum possible number of such primes p and a number 
bigger than any of the k(p)log(p).

For the terms with k(p) = 1, we have at most

     sum(log(p): p <= 2n/3) = theta(2n/3) < (4n/3) log(2)

by what we proved way back when.

Putting together what we've got so far gives us

     log N < (4n/3) log(2) + sqrt(2n) log(2n).

Time for another breather before we close in for the kill.

Looking back at the definition of N, we have

     2^(2n) = 2 + C(2n, 1) + C(2n, 2) + ... + C(2n, 2n-1)

(Binomial Theorem with first and last terms combined).

This is a sum of 2n terms, the largest of which is C(2n, n) or N. So

     2^(2n) < 2nN
     2n log(2) <  log(2n) + log(N)
               <= log(2n) + (4n/3) log(2) + sqrt(2n) log(2n)

by what we proved just before the breather.

Now for large values of n, the only term that counts on the right side 
is the 4n/3 log(2), which is smaller than the 2n log(2). So what we're 
going to do is figure out how big n needs to be to make this 
inequality false, and then just prove the postulate directly for 
smaller values of n. Take n >= 2^9 and note that 

     log(2n) = log(2^10) = 10 log(2)

Divide the inequality by log(2) to get

     2^10 < 10 + 2^10(2/3) + (2^5) 10


     2^10 (1 - 2/3) < 10 (2^5 + 1)

     2^10 (1/3) < 10 (2^5 + 1)

     2^10 < 30 (2^5 + 1) < 31 (2^5 + 1) = (2^5 - 1) (2^5 + 1)

                                        = 2^10 - 1

which is false!

So the assumptions that Bertrand's Postulate is false for n and that 
n >= 2^9 lead to a contradiction.  All that remains is to verify the 
postulate for n < 2^9 = 512.

Here we can just look at the sequence of primes

     2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631

each of which is less that twice the one before.

- Doctor Wilkinson, The Math Forum   
Associated Topics:
High School Discrete Mathematics
High School Number Theory

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