Associated Topics || Dr. Math Home || Search Dr. Math

### Consecutive Integers

```
Date: 02/10/99 at 13:11:27
From: Lori
Subject: Consecutive integers

I am trying to help my freshman son with his homework, but this one has
me stumped. I've been away from school for too long! Here's the
problem: Determine whether the product of four consecutive integers can
be a perfect square. Justify your conclusion.

Lori
```

```
Date: 02/10/99 at 17:35:34
From: Doctor Wilkinson
Subject: Re: Consecutive integers

Integers that differ by 3 or less cannot have any common factors other
than 2 and 3. So the product of two consecutive integers cannot have
any common factors with the product of the next two consecutive
integers other than 2, 3, and 6.

If the product of four consecutive positive integers is a square, call
the product of the first two a and the product of the second two b.
Then the only possibilities are that a and b are both squares, both
twice a square, both three times a square, or both six times a square.

Since any two consecutive positive integers are relatively prime, the
product of two consecutive integers can be a square only if both
actors are squares, and this is impossible. The product can be twice
a square only if  one is a square and the other is twice a square. If
the same is true of the next two integers, then you have two squares
that differ by at most 3. The only possibility is 1 and 4, and that
doesn't work: 1 * 2 * 3 * 4 is not a square.

Similar arguments will dispose of the remaining cases.

So far I have not been able to come up with anything simpler, so I
think probably the trick question is what was intended.

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

```
Date: 09/27/2001 at 21:51:48
From: Stephanie Schreibe
Subject: Proofs

I cannot figure out how to prove that any product of four consecutive
integers is equal to one less than a perfect square. I'm trying to
help a friend, and no one can remember any theorems that might be
useful. Any ideas?

Thank you,
Stephanie
```

```
Date: 09/28/2001 at 04:10:49
From: Doctor Floor
Subject: Re: Proofs

Hi, Stephanie,

Thanks for writing.

As a start, let us write out a product of four consecutive integers,
and expand it:

(n-1)n(n+1)(n+2) = (n-1)(n+1) * n(n+2)
= (n^2 - 1) (n^2 + 2n)
= n^4 + 2n^3 - n^2 - 2n

In order to prove your statement, we should check whether
n^4 + 2n^3 - n^2 - 2n + 1 is a square. If that is the case, then we
should have that n^4 + 2n^3 - n^2 - 2n + 1 is equal to:

(n^2 + an + b)(n^2 + an + b) = n^4 + 2an^3 + (a^2 + 2b)n^2 + 2abn + b^2

from the final term we conclude that b = 1 or b = -1.

Since the terms with n^3 and with n are opposites (2n^3 and -2n), we
see that b must be equal to -1. And from the term with n^3 we see that
a must be equal to 1. Inserting these values gives:

(n^2 + n - 1)(n^2 + n - 1) = n^4 + 2n^3 - n^2 - 2n + 1

And that is correct.

So we can conclude that (n-1)n(n+1)(n+2) is one less than
(n^2 + n - 1)^2, proving your question.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory