Question

If $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$, where

• A. $n\in N$.
• B. $n$ is odd natural number.
• C. $n$ is even natural number.
• D. $n$ is not a composite number.

Answer 1

The correct option is A $n\in N$.

$n^p-n$ is divisible by $p$ for $n=1$
Let, $k^p-k=p\lambda$, where $k\in N$ and $\lambda \in Z$

Now, ${(k+1)}^p-(k+1)=k^p{+}^p{\text{C}}_1{k}^{p-1}+.........{+}^p{\text{C}}_p-1-k$

$(k^p-k)+({}^p{\text{C}}_1{k}^{p-1}+...{+}^p{\text{C}}_{p-1}k)$

$=p\lambda +({}^p{\text{C}}_1{k}^{p-1}+...{+}^p{\text{C}}_{p-1}k)$
which is divisible by $p$
Reason :${}^p{\text{C}}_f$ is always divisible by by p when $f\ne p,0$ because p is a prime number as the denominator will not have any factors or multiples of p.

So by mathematical induction, $n^p-n$ is divisible by $p$ for all $n\in N$