Prove that for an integer $x \ge 7$, it follows that $x\# > x^2+x$

Question

Is the following argument sufficient to show that for an integer $x \ge 7, x\# > x^2 + x$.

Please let me know if I made a mistake or if there is a more straight forward way to make the same argument.

Let:

• $p_n$ be the $n$th prime
• $p\#$ be the primorial for $p$

Here's the argument by induction:

(1) Base case: $p_4=7$

For $7 \le x < 14, 7\# = 210 > x^2+x$ since:

$$7^2 + 7 < 8^2 + 8 < 9^2 + 9 < 10^2 + 10 < 11^2 + 11 < 12^2 + 12 < 13^2 + 13 = 182 < 210$$

(2) Assume up to some prime $p_n \ge 7$ that for $p_n \le x < 2p_n, p_n\# > x^2+x$.

(3) From Bertrand's Postulate, $p_n < p_{n+1} < 2p_n$

(4) $p_{n+1}\# > p_{n+1}[(2p_n-1)^2 + 2p_n-1] = p_{n+1}(2p_n)^2 - 2p_{n+1}p_n$

(5) Since $p_{n+1} \ge 11$, it follows from Bertrand's Postulate:

$$p_{n+1}(2p_n)^2 - 2p_{n+1}p_n > p_{n+1}(p_{n+1})^2 - 2p_{n+1}p_n > 9(p_{n+1})^2$$

(6) It follows:

$$p_{n+1}\# > (3p_{n+1})^2 > (2p_{n+1})^2 - 2p_{n+1} = (2p_{n+1} - 1)^2 + 2p_{n+1} - 1$$

(7) For any integer $x \ge 7$, let $p_n$ be the highest prime less than or equal to $x$.

(8) If $x$ is prime, from the above, $x\# > x^2 + x$

(9) If $x$ is not prime, from Bertrand's Postulate, it follows:

$$x\# = p_n\# > (2p_n-1)^2 + 2p_n - 1 \ge x^2 + x$$

Answer 1

What you've done looks correct. However, one small thing to note in your step $(1)$ base case, you only need to show $10^2 + 10 = 110 \lt 210$ since $11$ is prime and, as such, will be handled later by induction as $p_5$.

Also, I believe a somewhat simpler way to proceed after your step $(5)$ is to then show, for all $p_{n+1} \le x \lt 2p_{n+1}$, that

$$\begin{equation}\begin{aligned} p_{n+1}\# & \gt 9p_{n+1}^2 \\ & = 4p_{n+1}^2 + 5p_{n+1}^2 \\ & \gt (2p_{n+1})^2 + 2p_{n+1} \\ & \gt x^2 + x \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

As you stated in your step $(3)$, Bertrand's postulate shows there's a prime $p_{n+1} \lt p_{n+2} \lt 2p_{n+1}$. Thus, for all $p_{n+1} \le x \lt p_{n+2}$, you have $x\# = p_{n+1}\#$, with \eqref{eq1A} showing

$$x\# \gt x^2 + x \tag{2}\label{eq2A}$$

Answer 2

As an alternative, let's suppose we've verified the inequality for all integers up to $24$ (a simple extension of the OP's argument taking things up to $13$, since $2\cdot3\cdot5\cdot7\cdot11=2310\gt24^2+24$), and let's invoke Bertrand's Postulate in the equivalent form that for any real number $u\gt1$ there is a prime satisfying $u\lt p\lt2u$.

This version of BP allows for the following conclusion: If $x\gt24$, then there exist prime numbers $p$, $q$, and $r$ such that

$$2\lt3\lt{x\over8}\lt p\lt{x\over4}\lt q\lt{x\over2}\lt r\lt x$$

It follows that

$$\#x\ge6pqr\gt6\cdot{x\over8}\cdot{x\over4}\cdot{x\over2}={3x^3\over32}$$

and it's easy to see that

$${3x^3\over32}\gt x^2+x$$

if $x\gt24$ (in fact, if $x\ge12$).