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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

I have only been able to derive a proposed solution to the first part; although I confess that I am not entirely certain of its accuracy.

**PROPOSED SOLUTION TO FIRST PART**

Let the common difference of the sequence be r, so that x = y-r and z = y+r; so that:

y^4 + (y-r)^4 = (y+r)^4 - n

Or, y(8*r^3 + 8*y^2*r - y^3) = n --------(#)

Or, n+y^4 = 8*P*r, where P = r^2*y+ y^3.........(##)

Hence, from (#) and (##):

n must be divisible by y, and:

(n+y^4) must be divisible by 8r.

*Edited on ***August 5, 2006, 12:11 pm**