Natural Numbers Problem

Date: 12 May 1995 05:11:20 -0400
From: Dana Murray
Subject: Question 5

Good morning Dr Math

Thank you for the hints with question 4.

How would you prove that one added to the product of four consecutive
natural numbers is a perfect square?

Thank you

Date: 16 May 1995 22:11:17 -0400
From: Dr. Sydney
Subject: Re: Question 5

Hello again, Dana.

Let's attack this problem by writing everything out in some helpful
notation.  For a general case, we can write the product of four consecutive
natural numbers as n(n+1)(n+2)(n+3), where n is a natural number.  
If we were looking at the product 1*2*3*4, we would be looking at 
the n=1 case. If we were looking at the product 8*9*10*11, we would 
be looking at the n=8 case.  Do you get the idea?

So, our problem is to show that for any natural number, n, the number:

n(n+1)(n+2)(n+3) + 1

is a perfect square.  

Again, let me give you a hint, and you can then see if you can take it from
there, okay?  Simplify the number, n(n+1)(n+2)(n+3) + 1, by multiplying
through on the first term, combining like terms, etc. until you get a fourth
degree polynomial.  Can you factor this polynomial?  The proof will be
complete if you can factor the polynomial you get into a polynomial times
itself.  

If you can't seem to see what your fourth degree polynomial would factor
into, try this to figure it out.  Write out a chart with 3 columns.  Let the
number n be in the first column, the number n(n+1)(n+2)(n+3) be in the
second column (note don't multiply through, you'll be looking for a pattern
here that is easier to see if you write this as a product of 4 numbers), and
the number sqrt [n(n+1)(n+2)(n+3) + 1] (here sqrt means square root) be in
the third column.  Then the beginning of your chart will look like this (you
can verify this by calculation...)

1   1*2*3*4   5
2   2*3*4*5   11
3   3*4*5*6   19
4   4*5*6*7   29
.      .       .
.      .       .
.      .       .

Can you see any relations between the second and third columns? If you 
can, you come up with a relation between the first and second columns, 
then you won'tneed to factor that fourth degree polynomial.  Do you 
see how?

I'll let you work on this from here.  If any of this seems confusing or
unclear, write back, and I'll try and help.  Good luck!  Neat problem, too.

--Sydney, "dr. math"