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(C) Given $a^{n} + b^{n} = c^{n} + d^{n}$ for all $n \in \mathbb{N}$.For $n = 1$, we have $a + b = c + d$.For $n = 2$, we have $a^{2} + b^{2} = c^{2}

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$a + b = c + d$ and $a^{2} + b^{2} = c^{2} + d^{2}$

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(C) Given $a^{n} + b^{n} = c^{n} + d^{n}$ for all $n \in \mathbb{N}$.For $n = 1$, we have $a + b = c + d$.For $n = 2$, we have $a^{2} + b^{2} = c^{2} + d^{2}$.From $a + b = c + d$, we get $a - c = d - b$.From $a^{2} + b^{2} = c^{2} + d^{2}$, we get $a^{2} - c^{2} = d^{2} - b^{2}$, which implies $(a - c)(a + c) = (d - b)(d + b)$.If $a 
eq c$, then $a + c = d + b$. Since $a - c = d - b$, adding these gives $2a = 2d \Rightarrow a = d$, which implies $b = c$.If $a = c$, then $b = d$.In both cases, the set of values ${a, b}$ is the same as ${c, d}$.Option $(C)$ provides the conditions $a + b = c + d$ and $a^{2} + b^{2} = c^{2} + d^{2}$, which are necessary and sufficient for the equality to hold for all $n$.

Let $a, b, c$ and $d$ be any four real numbers. Then $a^{n} + b^{n} = c^{n} + d^{n}$ holds for any natural number $n$ if:

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