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
n=k(k+1)(k+2).
Thus, we are looking for s>k such that
s2-1|k(k+1)(k+2).
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
s2-1=k(k+2),
which divides
n=k(k+1)(k+2).
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Tags: Divisibility, Math, Monday Math, Number Theory
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.