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

Well, lets say that the three numbers are a, a+1, and a+2.

Then the product would be a^3 + 3a^2 + 2a, and the conjecture made in the problem is that a^3 + 3a^2 + 2a != n^x, where n and x are both integers and n is > 1.

Hmmm.. well I dont quite know how to continue with the proof from here... but I think the start is solid.  I will have to make a computer program to take some numbers and check the outputs... interesting.

Posted by Dan on 2005-09-05 21:37:02