Show that the product of three consecutive positive integers cannot be the n-th power of an integer, for any integer n>1.

Let our string of consec positive integers be (a-1), a and (a+1).

X= (a-1)a(a+1)=(a^3-a)=a(a^2-1)

We will prove by contradiction.  Say

a(a^2-1) = b^n           for some b, and n>1

(a^2-1) = (b^n)/a

We know the LHS is the product of 2 positive intergers and is thus a positive interger.  Therefore the RHS is a positive interger.

(b^n)/a =(b/a)b^(n-1)    Where b^(n-1) >=0 and an interger

Therefore (b/a) is a positive interger.  So b must either equal a or be a muliple of a.

b=ca for some interger c>=1

(a^2-1) = (b^n)/a and substituting for b

(a^2-1) = (ca)^n/a

(a^2-1) = (c^n)a^n-1

Which is not True !

Posted by Christie on 2005-09-20 20:58:36