Raja Oktovin wrote: Let $ a,b,c$ be non-zero real numbers satisfying \[ \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1}{a + b + c}.\] Find all integers $ n$ such that \[ \dfrac{1}{a^n} + \dfrac{1}{b^n} + \dfrac{1}{c^n} = \dfrac{1}{a^n + b^n + c^n}.\] I think it is true for all $ n = odd$ numbers....

Just a hint

Dimitris X wrote: I think it is true for all $ n = odd$ numbers.... Dimitris X is right : $ \frac 1a+\frac 1b+\frac 1c=\frac 1{a+b+c}$ $ \iff$ $ (ab+bc+ca)(a+b+c)=abc$ with $ abc\ne 0$ $ \iff$ $ (a+b)(b+c)(c+a)=0$ and $ abc\ne 0$ And so : $ (a+b)(b+c)(c+a)=0$ and $ abc\ne 0$ and $ n$ odd $ \implies$ $ (a^n+b^n)(b^n+c^n)(c^n+a^n)=0$ and so the same property for $ a^n,b^n,c^n$ And obviously, $ n=2p$ and $ abc\ne 0$ $ \implies$ $ (a^{2p}+b^{2p})(b^{2p}+c^{2p})(c^{2p}+a^{2p})\ne 0$

$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ $ \frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}$ $ (a+b+c)(ab+bc+ca)=abc$ $ a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc=abc$ $ a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+2abc=0$ $ (a+b)(b+c)(c+a)=0$ Then at least 1 of $ a+b,b+c,c+a$ is equal to $ 0$, WLOG : $ a+b=0$ ,$ b+c\ne c+a \ne 0$ , so $ a=-b$ $ \rightarrow$ $ a^n=-b^n$ (if $ n$ is odd) ,so $ a^n+b^n=0$ , so $ (a^n+b^n)(b^n+c^n)(c^n+a^n)=0$ and it will lead to $ \frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$ But if $ n$ is even,and because $ a,b,c$ is not equal to $ 0$ , then $ a^n+b^n\ne0$ ,, $ b^n+c^n\ne0$ and $ c^n+a^n\ne 0$ , so $ (a^n+b^n)(b^n+c^n)(c^n+a^n)\ne0$ So the solution are $ n=2k-1$ where $ k\in\mathbb{N}$