Let $p$ be a prime number and $k$ be a positive integer. If $p$ divides $k^2$, then which of these is definitely divisible by $p$?

$\frac{k}{2}$

$k$

$7k$

$k^3$

The value of $\frac{a+\sqrt{a^2-b^2}}{a-\sqrt{a^2-b^2}}+\frac{a-\sqrt{a^2-b^2}}{a+\sqrt{a^2-b^2}}$ is

To develop the proof of $\sqrt{5}$ be irrational.

Step I: Let $\sqrt{5}=\frac{a}{b},$ where $a$ and $b$ are _____; $b\ne 0$, $a$ and $b$ are _____.

Step II: On squaring both sides, we get
 $5=\frac{a^2}{b^2}\Rightarrow a^2=$_____.

$\Rightarrow a^2$ is divisible by _____.

$\Rightarrow a$ is divisible by _____.

Step III: Let $a=5\mathrm{k}$, $\mathrm{k}\in$_____.

On putting $a=5k$ in $\left(1\right)$, we get

_____$=5b^2\Rightarrow b^2=$_____.

$\Rightarrow b^2$ is divisible by _____.

$\Rightarrow b$ is  divisible by _____.

Hence, $a$ and $b$ have a common factor _____.

$\therefore$ Our supposition is wrong.

Hence, $\sqrt{5}$ is _____.