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The original problem:
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.
Solution.
I Algebraically, n^4-1=x(x-d)(x+d) with (x-d)=a, x=b, (x+d)=c.
II Set d, the difference constant, at (x-2)
III Then n^4-1=2*x*(2x-2)=4x^2-4x
IV 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.
Generalisation
1. That there are infinitely many solutions to the above equation for every positive integer k:
Generalisation to n^k-1: If we set n=y^4, then (y^k)^4-1 and the same reasoning 1-4 applies immediately for any k we choose.
2. Generalisation where k is even, say 2y:
i.(2a-1)^2=n^(2y)=(n^y)^2, 2a-1=(n^y), solves for all odd n, and clearly does NOT solve for any even n.
ii. If n is even, solutions still exist, if we are prepared to include negative integers as {a,b}; writing x(x-d)(x+d) as x(x^2-d^2) and setting x=-1 produces (n^y)^2=d^2, which solves easily for all n.
iii. Last but not least, there are infinite solutions for even n where k=2; then (2y)^2-1=x(x^2-d^2); (e.g) x=(d^2-1) or x=(d+1), as follows: (a) (n^m)^2-1 = x(x-(x-2))(x+(x-2)):x=1/2(1-n^m), x=1/2(n^m+1) using the first substitution (and n must be odd)
(b) (n^m)^2-1 = (d^2-1)((d^2-1)-d)((d^2-1)+d), n^m=(d-1)^2-1 using the second substitution. m=1 produces {3,8,15,24...}, m>1 has only one solution, {m=3, n=2, d=3} (Catalan)
(c) 8(n^m)^2=(4d+3)^2+7 using the third substitution, x=d+1:(a Pellian, with{X=-1, Y=1}{X=1,Y=1}
P,Q,R,S}={3,1,8,3})
We are looking for X a perfect power, m (but note that if X=1, then d is negative);
True if m=1 for X={1,4,23,134..}
True if m=2, for X={1,4(when d is 8)}
and for all m thereafter where X=1 (but the above note applies)
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