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Consecutive Integer Proof

Date: 09/16/2004 at 11:52:19
From: Mam
Subject: showing that five consecutives integers cannot be a square

Is it possible to prove that the product of five consecutive integers
cannot be a perfect square?

Thanks a lot.



Date: 09/16/2004 at 16:08:13
From: Doctor Vogler
Subject: Re: showing that five consecutives integers cannot be a square

Hi Mam,

Well, it *can* be zero, but I think that's the only perfect square it
can be.  So can we prove it?

That's a tough question!  I had to think about that for a while.  
Here's what I got:

So you have

  n * (n+1) * (n+2) * (n+3) * (n+4)

is a perfect square.  Now suppose that p is some prime number

  p >= 5

and p divides one of those numbers (that is, p divides n+i for some i
from 0 to 4 inclusive).  Then p cannot divide any of the other numbers
(why?) and therefore the largest power of p that divides n+i must be
an EVEN power.

In other words, if you divide out squares of primes from each of the
numbers n+i, then you can only be left with 2's and 3's.  So now let's
look at the 2's and 3's (separately).

Let's write a 2 for those n+i which are divisible by an odd number of
2's (like 2, 6, 8, 10, 32, and not 4, 12, or 64) and a 1 for those n+i
which are divisible by an even number of 2's.  For example, n = 4
would change

  4 * 5 * 6 * 7 * 8

into

  1 * 1 * 2 * 1 * 2.

Since this product is a square, we must have an even number of 2's.  
But if we have four of them, then two are adjacent, implying two even
numbers that differ by 1 (n and n+1 can't both be even).  So there are
either two 2's or none.  In fact, it's impossible that there be none,
since this would mean that every even number from n to n+4 is
divisible by 4, and that can't happen.  So there are exactly two 2's.
Then they will be in the order

  2 * 1 * 1 * 1 * 2

(as in 2, 3, 4, 5, 6) or as

  1 * 1 * 2 * 1 * 2
  1 * 2 * 1 * 2 * 1
  2 * 1 * 2 * 1 * 1

(as in 4, 5, 6, 7, 8 and 5, 6, 7, 8, 9 and 8, 9, 10, 11, 12).

Similarly, when we look at 3's, there are either two 3's or none, and
if there are two 3's, then they must have two 1's in between them, so
we have one of the three cases

  1 * 1 * 1 * 1 * 1
  3 * 1 * 1 * 3 * 1
  1 * 3 * 1 * 1 * 3

(as in 7, 8, 9, 10, 11 and 3, 4, 5, 6, 7 and 2, 3, 4, 5, 6).

If we put these together, we end up with twelve possibilities. 
Dividing n, n+1, n+2, n+3, n+4 by squares, we get one of the following
sequences of numbers left over:

  2 * 1 * 1 * 1 * 2
  1 * 1 * 2 * 1 * 2
  1 * 2 * 1 * 2 * 1
  2 * 1 * 2 * 1 * 1
  6 * 1 * 1 * 3 * 2
  3 * 1 * 2 * 3 * 2
  3 * 2 * 1 * 6 * 1
  6 * 1 * 2 * 3 * 1
  2 * 3 * 1 * 1 * 6
  1 * 3 * 2 * 1 * 6
  1 * 6 * 1 * 2 * 3
  2 * 3 * 2 * 1 * 3

Now we should notice that in each case we can either find two ones,
meaning that n+i and n+j are both squares, or we have two twos that
are two apart, meaning that n+i and n+j are each twice a square.  That
means that we have reduced the problem to looking for all integer
solutions to the equations

  a^2 - b^2 = 1
  a^2 - b^2 = 2
  a^2 - b^2 = 3
  2a^2 - 2b^2 = 2

I think you will quickly find that there aren't very many integer
solutions to any of these equations.  Then you just have to try each
of those (few) solutions in the correct places and see that the
product doesn't work.

For example,

  2 * 1 * 1 * 1 * 2

has n+1 = b^2 and n+2 = a^2, so that

  a^2 - b^2 = 1

or

  (a - b)(a + b) = 1

which means that either

  a - b = a + b = 1

or

  a - b = a + b = -1,

and in either case we have b = 0 and a^2 = 1, meaning that

  n+1 = b^2 = 0,
  n+2 = a^2 = 1,

so

  n = -1,

but then the product of the five numbers is

  -1 * 0 * 1 * 2 * 3 = 0.

You can check the other cases.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Number Theory

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