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### Factorials Can't Be Squares

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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.
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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

(2m+1)!/m!(m+1)!

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)

so

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

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

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

[n/p^m]

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
or
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

or

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
http://mathforum.org/dr.math/
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Associated Topics:
High School Discrete Mathematics
High School Number Theory

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