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Further Arithmetic Integers II (Posted on 2006-08-04) Difficulty: 4 of 5
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)

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Comments On Part Two | Comment 2 of 4 |

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
  Posted by K Sengupta on 2006-08-05 02:53:23

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