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:
N^{4} 1 = a*b*c
Prove that there are infinitely many solutions to the above equation.
1. Algebraically, n^41=x(xd)(x+d) with (xd)=a, x=b, (x+d)=c.
2. Set d, the difference constant, at (x2)
3. Then n^41=2*x*(2x2)=4x^24x
4. Adding 1 to both sides: 4x^24x+1=(2x1)^2=n^4; 2x1=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 20110418 13:43:46 