Problem Analysis #3

<|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 $a^4+b^4+c^4 = \boxed{89909683}$.