Consider five positive integers A < B < C < D < E in arithmetic sequence, and find all possible solutions of:
A⁴ + B⁴ + C⁴ + D⁴ = E⁴ - 143

Letting p equal the constant in the arithmetic progression, a quartic polynomial can be formed:
3A⁴ + (8p)A³ + (-12p²)A² + (-112p³)A + (143 - 158p⁴) = 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 2014-12-19 16:14:16