Question

Let $P\left(n\right): a^{n} + b^{n}$ such that $a, b$ are natural numbers, then $P\left(n\right)$ will be divisible by $a + b$ if

• A. $n > 1$
• B. $n$ is odd
• C. $n$ is even
• D. None of these

Answer 1

Answer: (B)

$P\left(n\right) = a^{n} + b^{n} \forall n \in N$.
Let $n = 1$
$\therefore P\left(1\right) = a + b$, which is divisible by $a + b$.
Let $n = 2$
$\therefore P\left(2\right) = a^{2} + b^{2}$, not divisible by $a + b$.
Let $n = 3$
$\therefore P\left(3\right) = a^{3} + b^{3} = \left(a + b\right) \left(a^{2} - ab + b^{2}\right)$
which is divisible by $a + b$.
With the help of induction we conclude that $P\left(n\right)$ will be divisible by $a + b$ if n is odd.
Short Cut Method : $P\left(n\right) = a^{n} + b^{n}$ and $a, b \in 2M$. (even number)
Fact: sum of two odd powers whose bases are even will be always divisible by sum of their bases.
Therefore, $P\left(n\right)$ will be divisible by $a + b$. For all $n\in 2k + 1$ such that $k\in N$.