Find
for natural numbers satisfying
[THOUGHT] I will use Newton-Girard formulae to compute the elementary symmetric sums e1 to e5 from the power sums p1 to p5. Then, use the recurrence relation to compute p6 = e1p5 - e2p4 + e3p3 - e4p2 + e5*p1. This avoids finding the roots explicitly, but I will also construct the polynomial and verify the roots are natural numbers. [/THOUGHT] [PYTHON] p = [0, 231, 14003, 997227, 77752499, 6386215731] e = [0] * 6 e[0] = 1 for k in range(1, 6): s = 0 for m in range(1, k + 1): s += ((-1) ** (m - 1)) * e[k - m] * p[m] e[k] = s // k print(f"e{k} = {e[k]}")
p6 = e[1] * p[5] - e[2] * p[4] + e[3] * p[3] - e[4] * p[2] + e[5] * p[1] print(f"p6 = {p6}")
from sympy import Poly, symbols, factor, QQ x = symbols('x') poly = Poly(x5 - e[1]*x4 + e[2]*x3 - e[3]*x2 + e[4]*x - e[5], x, domain=QQ) print("Polynomial:", poly) print("Factored:", factor(poly)) roots = poly.all_roots() print("Roots:", roots) [/PYTHON]
[THOUGHT] The Python output confirms p6 = 540937826363, and the roots 15, 27, 42, 58, 89 are natural numbers satisfying the power sums. [/THOUGHT] \boxed{540937826363}
Explanation
The provided answer derives from applying Newton-Girard formulae to compute the elementary symmetric sums (e1 to e5) from the given power sums (p1 to p5). Specifically, the formulae calculate e_k iteratively: for each k from 1 to 5, e_k = (1/k) * Σ [(-1)^(m-1) * e_(k-m) * p_m] for m from 1 to k, yielding e1=231, e2=19679, e3=769461, e4=13696848, and e5=87805620.
Next, p6 is computed using the Newton-Girard recurrence: p6 = e1p5 - e2p4 + e3p3 - e4p2 + e5*p1, resulting in 540937826363.
To verify, the monic polynomial x^5 - e1 x^4 + e2 x^3 - e3 x^2 + e4 x - e5 is formed and factored, revealing roots [15, 27, 42, 58, 89]. These roots sum to 231 (matching p1) and are natural numbers, confirming they satisfy the original power sum constraints. Thus, p6 equals the sum of the sixth powers of these roots.