Squares in an Infinite Factorial Series

Date: 11/23/2001 at 10:32:31
From: Gaurav Tekriwal
Subject: Squares in an Infinite Factorial Series

How many perfect squares appear among the following numbers?

   1!, 1!+2!, 1!+2!+3!,...1!+2!+3!+...n! 

My approach was looking for patterns or cycles as such, but I wasn't 
able to find any. Is there an interesting way to find the number of 
perfect squares?

I also think that the number would tend to infinity.
Also, if we bound the sequence to a limit, say k!, then is it possible 
to find the number of perfect squares?

Thanks,
GAuRAv

Date: 11/23/2001 at 11:07:25
From: Doctor Tom
Subject: Re: Squares in an Infinite Factorial Series

Hi Gaurav,

You didn't look too carefully:  1! + 2! + 3! = 1 + 2 + 6 = 9.

I'm not sure how to prove whether or not there are an infinite number, 
but there's at least one. 

- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/   

Date: 11/23/2001 at 12:42:31
From: Doctor Jubal
Subject: Re: Squares in an Infinite Factorial Series

Hi Gaurav,

The solution that Dr. Tom found, 9 = 3^2 = 1!+2!+3!, along with 
1 = 1^2 = 1!, are the only solutions. Here's how you can prove this is 
true.

Let's look at the next largest term in the sequence

  1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

Note the last digit of 33 is 3. Now, any factorial greater than 
4! = 24 is divisible by 10, because they have the form 1*2*3*4*5*(more 
stuff), and so are divisible by both 2 and 5. Thus, the last digit of 
all factorials beginning with 5! = 120, is 0. No matter how many times 
we add zero to 3, we still get 3, so the last digit in every one of 
the sums of the factorials you're interested in, beginning with 33, 
is 3.

However, no perfect square has 3 as its last digit. To prove this, 
observe that we can write any integer as n = 10a + b, where b is the 
last digit and a is the rest of the digits. Then

  n^2 = (10a + b)^2 = 100a^2 + 20ab + b^2

100a^2 and 20ab both are divisible by 10 and so have zero as the last 
digit, so the last digit of n^2 depends only on b^2. Thus, the last 
digit of any integer squared is the same as the square of the number's 
last digit.

Looking at the ten digits

   0^2 =  0
   1^2 =  1
   2^2 =  4
   3^2 =  9
   4^2 = 16
   5^2 = 25
   6^2 = 36
   7^2 = 49
   8^2 = 64
   9^2 = 81

None of them has 3 as the last digit of its square, so no number has 
three as the last digit of its square. All the factorial sums you're 
looking at that are greater than 9 end in three, so none of them can 
be a perfect square.

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/