Further Arithmetic Integers
K Sengupta asked for integer solutions
to x^4+y^4=z^4-64 with z>y>x>0 and in arithmetic sequence
My question is concerning the more general problem
x^4+y^4=z^4-n with z>y>x>0 and in arithmetic sequence
and for what values of n are there only two solutions
So far, I have not been able to derive any satisfactory method for solving the second part. However, I have noted a mathematical relationship which may lead to the solution.
Let the two triplets satisfying the conditions of the problem be given by (y-r, y, y+r) and (t-s, t, t+s).
Then, by conditions of the problem,
y(8*r^3 + 8*y^2*r - y^3)
= t(8*s^3 + 8*t^2*s - t^3) = n, ---------------------------------(##)
n must be separately divisible by y and t, so that:
(i) n is divisible by y*t/k where k is the g.c.d of y and t.
(ii) Also, (n+y^4) is divisible by 8r and (n+ t^4) is divisible by 8s.
So, relationships (i) and (ii) with regard to (##) may lead to a methodology culminating in a solution to the given problem.
Edited on August 5, 2006, 9:18 pm