<|channel|>commentary to=python code<|message|>s1 = 151 p3 = 957871 s2 = 5931 p2 = 10939 s3 = 67221 p1 = 151
p4 = s1p3 - s2p2 + s3*p1 p4
<|channel|>commentary to=python code<|message|># Let's recompute using Python s1 = 151 p3 = 957871 e2 = 5931 p2 = 10939 e3 = 67221 p1 = 151
first = s1 * p3 second = e2 * p2 third = e3 * p1
first, second, third, first - second + third
<|channel|>commentary to=python code<|message|>import sympy as sp sp.factorint(67221)
<|channel|>commentary to=python code<|message|>s1 = 151 e2 = 5931 p3 = 957871 s1**3
<|channel|>commentary to=python code<|message|>151*10939
<|channel|>commentary to=python code<|message|>s1 = 151 sum_cubes = 957871 e2 = 5931 s13, sum_cubes - s13 + 3s1e2
<|channel|>commentary to=python code<|message|>e1 = 151 p3 = 957871 e2 = 5931 p2 = 10939 e3 = 66221 p1 = 151
p4 = e1p3 - e2p2 + e3*p1 p4
<|channel|>analysis to=python code<|message|>151*10939
[THOUGHT] We have computed the symmetric sums:
- (e_1 = a+b+c = 151)
- (e_2 = ab+bc+ca = \frac{(a+b+c)^2 - (a^2+b^2+c^2)}{2} = \frac{151^2-10939}{2}=5931)
- (e_3 = abc = \frac{(a+b+c)(ab+bc+ca)-(a^3+b^3+c^3)}{3} = \frac{151\cdot5931-957871}{3}=66221)
Using Newton’s identities, the fourth power sum is (p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1), where (p_3 = a^3+b^3+c^3=957871), (p_2 = a^2+b^2+c^2=10939), and (p_1=e_1=151).
Thus (p_4 = 151\cdot957871 - 5931\cdot10939 + 66221\cdot151 = 89,909,683.)
So (a^4+b^4+c^4 = \boxed{89909683}).