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 Further Arithmetic Integers II (Posted on 2006-08-04)
in 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

 No Solution Yet Submitted by Daniel Rating: 5.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Proposed Method: Part I | Comment 1 of 4

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
 Posted by K Sengupta on 2006-08-05 02:42:57

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