Monday Math 137

Let n be the product of three consecutive positive integers. Show that there is an integer s, greater than the smallest of the consecutive integers, such that n is divisible by s2-1.

This is actually very simple, if one puts some thought into it. Let the consecutive integers be k, k+1, and k+2, so that
Thus, we are looking for s>k such that
Well, note that s2-1=(s-1)(s+1),
Thus, s2-1 is the product of two integers that differ by two. Note, then, that n has an obvious pair of factors that differ by two: k and k+2. Thus,

so s=k+1>k gives us an
which divides


Tags: , , ,

One Response to “Monday Math 137”

  1. Art Pasternak Says:

    If you make the integers (k-1), k and (k+1) it is even cleaner, since the product is k times (K^2 – 1) so it is obvious that s = k.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: