[THOUGHT] I will:

1. Store the given power sums p1..p5.
2. Use Newton's identities to compute the elementary symmetric sums e1..e5:
   • e1 = p1
   • p2 - e1p1 + 2e2 = 0, etc.
3. Then use the Newton recurrence for k>5: p6 = e1p5 - e2p4 + e3p3 - e4p2 + e5*p1
4. Print p6, which is a^6 + ... + e^6. [/THOUGHT] [PYTHON] from fractions import Fraction

Given power sums

p1 = 231 p2 = 14003 p3 = 997227 p4 = 77752499 p5 = 6386215731

Compute elementary symmetric sums using Newton's identities

e1 = p1

p2 - e1p1 + 2e2 = 0 -> e2

e2 = Fraction(e1*p1 - p2, 2)

p3 - e1p2 + e2p1 - 3*e3 = 0 -> e3

e3 = Fraction(p3 - e1p2 + e2p1, 3)

p4 - e1p3 + e2p2 - e3p1 + 4e4 = 0 -> e4

e4 = Fraction(e1p3 - e2p2 + e3*p1 - p4, 4)

p5 - e1p4 + e2p3 - e3p2 + e4p1 - 5*e5 = 0 -> e5

e5 = Fraction(p5 - e1p4 + e2p3 - e3p2 + e4p1, 5)

print("e1..e5:", e1, e2, e3, e4, e5)

Now compute p6 using recurrence for k>5:

p6 = e1p5 - e2p4 + e3p3 - e4p2 + e5*p1

from math import isfinite p6 = e1p5 - e2p4 + e3p3 - e4p2 + e5*p1 print("p6 as Fraction:", p6, "is integer?", p6.denominator == 1) print("p6 integer:", int(p6)) [/PYTHON]

[THOUGHT] Using Newton's identities on the given power sums, I computed the elementary symmetric sums and then applied the recurrence to obtain the 6th power sum, which matches the Python output. [ a^6 + b^6 + c^6 + d^6 + e^6 = 540937826363 ] \boxed{540937826363}