Proving a relation with induction I have a problem:
Let $p_n$ be the $n:th$ prime number ($p_1=2,  p_2=3,  p_3=5$ and so on). With induction, show that $p_{n+2}>3n$ for each integer $n\geq1$.
I can't figure this out because the primes are confusing me, making me unable to show the inductive step.
 A: For $n=1$ we observe $p_1=2, p_2=3, p_3=5$. Thus, $p_{1+2}=5>3$. So, let $n\in \mathbf{N}$ be such that the relation holds for $n$, then we show $p_{n+3} > 3(n+1)$. 
Now, $p_{n+3} \geq p_{n+2}+2 > 3n +2$, by the inductive hypothesis. So, $p_{n+3} \geq 3n+3 = 3(n+1)$. Since, $3(n+1)$ is composite, it cannot be prime. Hence, $p_{n+3} > 3(n+1)$.
So, the relation holds for all $n \in \mathbf{N}$.
My apologies for the blatantly incorrect statement I wrote before.
A: $p_{(n+1)+2}=p_{n+3}\geq p_{n+2}+2\geq3n+3$ now explain why $p_{n+3}\neq 3(n+1)$.
A: Proof by induction.
Let's start with the case $n=1$:
$p_{1+2}=p_3=5>3*1=3$
So the equality is true for $n=1$.
Now, we assume that the equality is true for the case $n-1$, that is,
$p_{n-1+2}=p_{n+1}>3(n-1)=3n-3$
and let's see that it is also true for n:
$p_{n+2}\geq p_{n+1}+2>3(n-1)+2=3n-1$; so $p_{n+2}\geq3n$
Now,you have to see that $p_{n+2}>3n$
A: For $n=1$ we can verify that $p_3=5>3\cdot1$
Suppose that the statement holds true for $n$ meaning that $p_{n+2}>3n$. We will show that
this holds for $n+1$.
 We have $p_{(n+1)+2}=p_{n+3}$. We already know from the induction step that $p_{n+2}>3n$  which means $p_{n+2}\geq 3n+1$.
So if $p_{n+3}\leq 3(n+1)=3n+3$ we have only one  possibility: $p_{n+3}=3n+2$   

(or $p_{n+3}=3n+1$ which means that $p_{n+2}$ and $p_{n+3}$ coincide .
  Also, if $p>3$ then $p$ cannot be  a multiple of $3$ so the case
  $p_{n+3}=3(n+1)$ is rejected)

But this means that $p_{n+2}$ and $p_{n+3}$ are consecutive primes which is only possible if the primes are $2,3$ (we are already above $5$)
All at the above step arrive at a contradiction and so we must have only one possibility: $p_{n+3}>3(n+1)$   
this proves the case for $n+1$ and so, for every $n$.
