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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, ---------------------------------(##)

so that:

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**