N is an odd positive integer > 2 and each of a, b and c is a positive integer. It is known that; a, b and c (in this order), with a < b < c, corresponds to three consecutive terms of an arithmetic sequence such that:
N4 -1 = a*b*c
Prove that there are infinitely many solutions to the above equation.
1. Algebraically, n^4-1=x(x-d)(x+d) with (x-d)=a, x=b, (x+d)=c.
2. Set d, the difference constant, at (x-2)
3. Then n^4-1=2*x*(2x-2)=4x^2-4x
4. Adding 1 to both sides: 4x^2-4x+1=(2x-1)^2=n^4; 2x-1=n^2; since n just has to be odd to meet this criterion, it follows that there are infinitely many solutions.
Most of Kth power and Consecutive Concern - Generalisation should now be straightforward.
Edited on April 18, 2011, 2:06 pm
Posted by broll
on 2011-04-18 13:43:46