Prove that ## p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2} ##

[Math100]

Homework Statement

Let $p_{n}$ denote the nth prime number. For $n\geq 3$, prove that $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$.
[Hint: Note that $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}$.]

Relevant Equations

None.

Proof:

Let $n\geq 3$ be an integer.
Applying the Bertrand's Postulate produces:
$p_{n+1}<2p_{n}$.
Then $p_{n+3}<2p_{n+2}$.
Thus $p_{n+3}^{2}<4p_{n+2}^{2}<4p_{n+2} (2p_{n+1})=8p_{n+2} p_{n+1}$.
Now we consider two cases.
Case #1: Suppose $n=3$.
Then $p_{6}^{2}=169<p_{3} p_{4} p_{5}=385$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=3$.
Case #2: Suppose $n=4$.
Then $p_{7}^{2}=289<p_{4} p_{5} p_{6}=1001$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=4$.
Therefore, $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n\geq 3$.

 

[Mark44]

Homework Statement:: Let $p_{n}$ denote the nth prime number. For $n\geq 3$, prove that $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$.
[Hint: Note that $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}$.]
Relevant Equations:: None.

Proof:

Let $n\geq 3$ be an integer.
Applying the Bertrand's Postulate produces:
$p_{n+1}<2p_{n}$.
Then $p_{n+3}<2p_{n+2}$.
Thus $p_{n+3}^{2}<4p_{n+2}^{2}<4p_{n+2} (2p_{n+1})=8p_{n+2} p_{n+1}$.

So you have applied the hint, but the work below doesn't take things any further.

Now we consider two cases.
Case #1: Suppose $n=3$.
Then $p_{6}^{2}=169<p_{3} p_{4} p_{5}=385$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=3$.
Case #2: Suppose $n=4$.
Then $p_{7}^{2}=289<p_{4} p_{5} p_{6}=1001$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=4$.
Therefore, $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n\geq 3$.

This won't fly. You can't jump from establishing the equation for $p_6^2$ and $p_7^2$, and then make the leap ("Therefore") to $p_n^2$.

 

[Math100]

What should I do then?

 

[Mark44]

You need to continue what you were doing with the hint.

 

[Math100]

Let $n\geq 3$ be an integer.
Applying the Bertrand's Postulate produces:
$p_{n+1}<2p_{n}$.
Now we consider two cases.
Case #1: Suppose $n=3$.
Then $p_{6}^{2}=169<p_{3} p_{4} p_{5}=385$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=3$.
Case #2: Suppose $n=4$.
Then $p_{7}^{2}=289<p_{4} p_{5} p_{6}=1001$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=4$.
Assume $p_{n+1}<2p_{n}$ for $n\geq 5$.
Then $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}<11p_{n+1} p_{n+2}=p_{5} p_{n+1} p_{n+2}\leq p_{n} p_{n+1} p_{n+2}$.
Note that $p_{n}\geq 11$ for $n\geq 5$.
Thus $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}<p_{n} p_{n+1} p_{n+2}$.
Therefore, $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n\geq 3$.

 

[Mark44]

Let $n\geq 3$ be an integer.
Applying the Bertrand's Postulate produces:
$p_{n+1}<2p_{n}$.
Now we consider two cases.
Case #1: Suppose $n=3$.
Then $p_{6}^{2}=169<p_{3} p_{4} p_{5}=385$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=3$.
Case #2: Suppose $n=4$.
Then $p_{7}^{2}=289<p_{4} p_{5} p_{6}=1001$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=4$.
Assume $p_{n+1}<2p_{n}$ for $n\geq 5$.
Then $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}<11p_{n+1} p_{n+2}=p_{5} p_{n+1} p_{n+2}\leq p_{n} p_{n+1} p_{n+2}$.
Note that $p_{n}\geq 11$ for $n\geq 5$.
Thus $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}<p_{n} p_{n+1} p_{n+2}$.
Therefore, $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n\geq 3$.

This is much better, although there is still a lot of stuff here that doesn't need to be included. E.g. in the 4th line under Case#2, you say "Assume $p_{n+1}<2p_{n}$ for $n\geq 5$."
You don't need to include Bertrand's Postulate again.

You also don't need Case 2, where n = 4.
Edit: It turns out that you do need this case.
If n = 4, the hint doesn't apply, but $289 = 17^2 = p_{4 + 3}^2 = p_7^2 < p_4p_5p_6 = 7*11*13 = 1001$.
So the statement is proved for n = 4.

If n = 3, you're dealing with $p_6, p_5,$ and $p_4$, which are 13, 11, and 7, respectively.
$169 = p_6^2 < 8p_1p_2 = 8\cdot11\cdot7 = 615$, so the statement is true for n = 3

If n > 3 n > 4 the smallest prime of the three primes on the right side of the statement to be proved is always $\ge 11$, so $p_{n+3}^3 < 8p_{n+1}p_{n + 2} < p_np_{n+1}p_{n + 2}$.

 

[fresh_42]

If n > 3, the smallest prime of the three primes on the right side of the statement to be proved is always $\ge 11$, so $p_{n+3}^3 < 8p_{n+1}p_{n + 2} < p_np_{n+1}p_{n + 2}$.

$p_4=7 < 8.$

 

[Mark44]

$p_4=7 < 8.$

If n = 4, then $p_{n + 3} = p_7$, which is 11, and 8 < 11.

 

[fresh_42]

If n = 4, then $p_{n + 3} = p_7$, which is 11, and 8 < 11.

$p_n$ is relevant, not $p_{n+3}$ (compare post #1 and hint therein).

 

[Mark44]

$p_n$ is relevant, not $p_{n+3}$ (compare post #1 and hint therein).

Here's the problem statement and hint, from post #1.

Let $p_{n}$ denote the nth prime number. For $n\geq 3$, prove that $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$.
[Hint: Note that $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}$.]

My work for $p_4$ had on off-by-one error, so this prime needs its own case, contrary to what I said earlier.
$289 = p_7^2 < 8*p_5*p_6$ per the hint. While it's true that $8 > p_4 = 7$, it is nevertheless true that $289 < p_4p_5p_6 = 7*11*13 = 1001$
For n = 5, we have $361 = 19^2 = p_8^2 < 8p_6p_7 < p_5p_6p_7 = 11*13*17 = 2431$
For n > 5, the smallest prime in the product of the three primes, $p_n$ is at least 13, which is larger than 8.
So for any integer $n \ge 3$, $p_{n + 3}^2 < p_np_{n+1}p_{n+2}$.

 

[Prof B]

Let $n\geq 3$ be an integer.
Applying the Bertrand's Postulate produces:
$p_{n+1}<2p_{n}$.

That statement is not relevant at this point.

Now we consider two cases.
Case #1: Suppose $n=3$.
Then $p_{6}^{2}=169<p_{3} p_{4} p_{5}=385$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=3$.
Case #2: Suppose $n=4$.
Then $p_{7}^{2}=289<p_{4} p_{5} p_{6}=1001$.
Thus $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n=4$.
Assume $p_{n+1}<2p_{n}$ for $n\geq 5$.

Two problems: 1. There is no reason to assume that, because it is implied by Bertrand's postulate. 2. You are about to use Bertrand's postulate twice. You have to state exactly how you are using it. Merely restating the postulate does not do the job.

Then $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}<11p_{n+1} p_{n+2}=p_{5} p_{n+1} p_{n+2}\leq p_{n} p_{n+1} p_{n+2}$.

You have to justify the first two inequalities. Use Bertrand's postulate to do so.

Note that $p_{n}\geq 11$ for $n\geq 5$.
Thus $p_{n+3}^{2}<4p_{n+2}^{2}<8p_{n+1} p_{n+2}<p_{n} p_{n+1} p_{n+2}$.

You've said this twice now. Once is usually enough.

Therefore, $p_{n+3}^{2}<p_{n} p_{n+1} p_{n+2}$ for $n\geq 3$.