Question

If $n$ is an odd positive integer, then $a^n+b^n$ is divisible by

• A. $a-b$
• B. $a+b$
• C. $a^2+b^2$
• D. none of these

Answer 1

The correct option is B $a+b$

Let, $P\left(n\right)=a^n+b^n$

$P\left(1\right)=a+b$, which is divisible by $a+b$

Now let $P\left(k\right)=a^k+b^k$ is divisible by $a+b$, where $k$ is an odd integer.

$\Rightarrow a^k+b^k=(a+b)f(a,b).........................\left(1\right)$

Now, $P(k+2)=a^{k+2}+b^{k+2}=a^2\left[\right(a+b\left)f\right(a,b)-b^k]+b^{k+2}$

$=a^{2}f(a,b)(a+b)-a^2{b}^k+b^{k+2}$ from $\left(1\right)$

$=a^{2}f(a,b)(a+b)-b^k(a^2-b^2)$

$=(a+b)\left[a^{2}f\right(a,b)-b^k(a-b\left)\right]$, which is divisible by $(a+b)$

Hence $a^n+b^n$ is divisible by $(a+b)$ for all odd positive integral $n$