Find
for natural numbers satisfying
[THOUGHT] I will:
- Use Newton's identities with the given power sums p1..p4 to solve for the elementary symmetric sums e1,e2,e3,e4 of a,b,c,d.
- Then use the recurrence for n>4: p_n = e1p_{n-1} - e2p_{n-2} + e3p_{n-3} - e4p_{n-4} to compute p5 = a^5 + b^5 + c^5 + d^5.
- Optionally verify that the resulting quartic has natural number roots. [/THOUGHT] [PYTHON] from sympy import symbols, Eq, solve
Given power sums
p1 = 151 p2 = 7989 p3 = 511201 p4 = 35858673
e1, e2, e3, e4 = symbols('e1 e2 e3 e4')
Newton's identities:
p1 - e1 = 0
eq1 = Eq(p1 - e1, 0)
p2 - e1p1 + 2e2 = 0
eq2 = Eq(p2 - e1p1 + 2e2, 0)
p3 - e1p2 + e2p1 - 3*e3 = 0
eq3 = Eq(p3 - e1p2 + e2p1 - 3*e3, 0)
p4 - e1p3 + e2p2 - e3p1 + 4e4 = 0
eq4 = Eq(p4 - e1p3 + e2p2 - e3p1 + 4e4, 0)
sol = solve([eq1, eq2, eq3, eq4], [e1, e2, e3, e4], dict=True) sol print(sol)
Extract the symmetric sums
e1_val = sol[0][e1] e2_val = sol[0][e2] e3_val = sol[0][e3] e4_val = sol[0][e4]
print("e1,e2,e3,e4:", e1_val, e2_val, e3_val, e4_val)
Compute p5 using recurrence: p5 = e1p4 - e2p3 + e3p2 - e4p1
p5 = e1_valp4 - e2_valp3 + e3_valp2 - e4_valp1 print("p5 =", p5)
Optionally verify roots are natural numbers
from sympy import Poly, factor, symbols x = symbols('x') poly = x4 - e1_val*x3 + e2_valx**2 - e3_valx + e4_val print("Polynomial:", poly) print("Factored:", factor(poly))
Evaluate roots numerically to confirm
roots = solve(poly, x) print("Roots:", roots)
[THOUGHT] Using Newton's identities and the given sums of powers, the computed value of (a^5 + b^5 + c^5 + d^5) from the valid Python output is 2624775001. [ \boxed{2624775001} ]