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 *s*^{2}-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

*n*=*k*(*k*+1)(*k*+2).

Thus, we are looking for *s*>*k* such that

*s*^{2}-1|*k*(*k*+1)(*k*+2).

Well, note that *s*^{2}-1=(*s*-1)(*s*+1),

Thus, *s*^{2}-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

*s*^{2}-1=*k*(*k*+2),

which divides

*n*=*k*(*k*+1)(*k*+2).

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Tags: Divisibility, Math, Monday Math, Number Theory

This entry was posted on September 27, 2010 at 12:03 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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November 1, 2010 at 8:04 am |

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.