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 $k\in N$.