Why is it true that if $p$ is prime and $n \in \mathbb N$ then $p^{n+1} \ne n$
$p$ is prime and $n \in \mathbb N$ then $p^{n+1} \ne n$
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prime-numbers
2 Answers
1
This is true because if $p$ is a prime number, then $p\geq 2$.
You can then easily prove by induction that
$$\forall n\in\mathbb N,\quad 2^{n+1}>n.$$
Because:
$2^{0+1}=2>0$.
$2^{1+1}=4>2$.
If $2^{n+1}>n$, than $2^{n+2}>2n\geq n+1$ if $n\geq 1$.
Plus, $p^{n+1}\geq 2^{n+1}$ which conclude the proof.
2
Proof (by contradiction): Assume that $p$ is prime and suppose $p^{n+1}=n$. Then $p^n\cdot p=n$ implying that $p^n$ divides $n$. This means that $p^n\leq n$. But since $p$ is prime, $p\geq 2$ and hence, $$p^n\geq 2^n>n.$$ We get a contradiction. Note that $2^n>n$ for all $n\in\Bbb N$ can be proved by using induction or simply by using the Bernoulli Inequality.