Too Large To Factor

Let $S_1 = a+b+c$ , $S_2 = ab+bc+ca$ , $S_3 = abc$ and $P_n = a^n + b^n + c^n$ , where $n$ is a natural number. Then we have:

$\begin{aligned} P_1 & = S_1 \\ P_2 & = S_1P_1 - 2S_2 \\ & = S_1^2 - 2S_2 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 \\ & = S_1^3 - 2S_1S_2 - S_1S_2 + 3S_3 \\ & = S_1^3 - 3S_1S_2 + 3S_3 \end{aligned}$

It is given that $P_3 = S_1^3$ , $\implies \color{#3D99F6}{S_1S_2 - S_3 = 0}$ .

Now, we have:

$\begin{aligned} P_5 & = S_1P_4 - S_2P_3+S_3P_2 \\ & = S_1(S_1P_3 - S_2P_2+S_3P_1) - S_2S_1^3 + S_3P_2 \\ & = S_1^5 - S_1S_2P_2+S_1^2S_3 - S_1^3S_2 + S_3P_2 \\ & = S_1^5 - (\color{#3D99F6}{S_1S_2-S_3})P_2 - S_1^2(\color{#3D99F6}{S_1S_2-S_3}) \\ & = S_1^5 - \color{#3D99F6}{0} - \color{#3D99F6}{0} \\ \implies a^5 + b^5 + c^5 & = \boxed{(a+b+c)^5} \end{aligned}$

As for the first equation, we have: $a^3+b^3+c^3=(a+b+c)^3 \Leftrightarrow (a+b)(b+c)(c+a)=0$

Let's consider one particular case where $a=-b$ , then:

$a^5+b^5+c^5 = a^5+(-b)^5+c^5=c^5$

and :

$(a+b+c)^5=[a+(-b)+c]^5=c^5$

which means that:

$(a^5+b^5+c^5)=(a+b+c)^5$ .

This can be generalised for any triplet of $(a,b,c)$ that satisfies $(a+b)(b+c)(c+a)=0$

True. If b=minus(c) then for any given a both the equations will be satisfied. basically it will be reduced to a^x on both the sides as long as x is an odd number. Did this in a hurry so I might be wrong.

I thought this was a logic problem and since since the hypothesis is false then the conclusion is true.I haven't worded it out in awhile.