Consider five positive integers A < B < C < D < E in arithmetic sequence, and find all possible solutions of:
A^{4} + B^{4} + C^{4} + D^{4} = E^{4}  143
Letting p equal the constant in the arithmetic progression, a quartic polynomial can be formed:
3A
^{4} + (8p)A
^{3} + (12p
^{2})A
^{2} + (112p
^{3})A + (143  158p
^{4}) = 0
Solving the quartic for p=1, the two real roots are (3, ≈66/485).
Iterating through different constants and values of A (using a computer program) found no other solutions.
As A and p (by inference) are given to be positive integers:
A = 3, B = 4, C = 5, D = 6, and E = 7.
Edited on December 19, 2014, 4:16 pm

Posted by Dej Mar
on 20141219 16:14:16 