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Everything 2014++

# Number Theory 2014++

## Prime product theorems

Let  $P:=\{2\ 3\ 5\ 7\ 11\ \ldots\}$  be the set of all primes. Let

$$\Pi_x := \{p\in P : p\le x\}$$

for every positive real number  $x$.  In note  Prime products--a modest beginning  I showed a simple proof of inequality

$$\Pi_n\ge n$$

for every natural number  $n$.  This is an infinitely weaker version of the well known results but it already shows that there are infinitely many primes, and in a simple way (only the original Euclid proof is still simpler while the given inequality seems to prove a little more). Anyway, I'd like to show stronger results along the similar line (while still a far cry from the ambitious classical known theorems).

THEOREM 0   $\forall_{n\ge 10}\ \,\Pi_{\frac n2}\gt n$,   where  $n$  is understood to be an arbitrary natural number.

PROOF  The theorem holds for  $n=10$.  Now let  $n>10$,  and--by induction--we assume that the theorem holds for  $n-1$,  i.e.  $\Pi_{\frac{n-1}2}>n-1$.  Then

$$\Pi_{\frac n2}\ge n$$

Thus all that we have to prove is that the equality is impossible. Indeed (arguing by a contradiction), assume that:

$$\Pi_{\frac n2} = n$$

Then  $n$  is a product of  $2$  and of an odd integer (i.e. a product of some odd primes). Thus  $n\equiv 2\mod 4$  hence  $m:=n-4\equiv 2\mod 4$.  It follows that  $\frac m2$  is an odd integer. Furthermore,  $\frac m2 = \frac {n-4}2 \ge \frac 72 > 1$.  Thus there exists an odd prime  $p|\frac m2$  hence  $p|m$.  However  $p$  does not divide  $4=n-m$  hence  $p$  does not divide  $n$.  This contradicts the equality  $n = \Pi_{\frac n2}$  (remember that  $p\le \frac m2<\frac n2$).  END OF PROOF

We can continue in the same vein:

THEOREM 1   $\forall_{n\ge 20}\ \,\Pi_{\frac n4}\gt n$,   where  $n$  is understood to be an arbitrary natural number.

PROOF  The theorem holds for  $n=20$. Let  $n>20$,  and--by induction--we assume that the theorem holds for  $n-1$,  i.e.  $\Pi_{\frac{n-1}4}>n-1$.  Then.

$$\Pi_{\frac n4}\ge n$$

Thus all that we have to prove is that the equality is impossible. Indeed (arguing by a contradiction), assume that:

$$\Pi_{\frac n4} = n$$

Thus  $n$  is a product of  $2$  and of an odd integer (a product of odd primes), i.e.  $n\equiv 2\mod 4$.  Let's consider two cases:

Case 1:  assume that  $m:=n-2$  is not a power of $2$, i.e. there exist a prime odd divisor  $p$  of  $m$.  But  $4|m$.  It follows that

$$p\le \frac m4 < \frac n4$$

which would mean that  $p|n = \Pi_{\frac n4}$.  This would mean that   $p|4=n-m$,  a contradiction. Thus this case is impossible.

Case 2:  assume that  $n-2 = 2^d$ for certain positive integer  $d$. Thus  $2^d = n-2\ge 19$,  i.e.  $d\ge 4$.  Define  $m:=n-6$.  Then

$$m=4\cdot(2^{d-2}-1)$$

is a product of  $4$  and of an odd integer  $s:=2^{d-2} - 1$,  where  $s\ge\lceil \frac{n-6}4\rceil \ge\lceil \frac{15}4\rceil =6$.  Thus exponent  $d-2$  is at least  $3$, i.e.  $d\ge 5$,  and  $s\ge 7$.

Let  $p$  be an arbitrary prime divisor of  $m$  (i.e. of  $s$).  Then  $p\le\frac m4 < \frac n4$  hence  $p|n$.  Thus  $p|n-m=6$  hence  $p=3$.  Since odd prime divisor  $p$  of  $m$  was arbitrary, it follows that  $m=4\cdot 3^c$  for certain positive integer  $c$. Thus the following equality holds:

$$2^\delta-1 = 3^c$$

where  $\delta := d-2$.  As can be seen at s.e.d.e., the only solutions  $(c\ \delta)$  of the above equation in non-negative integers are  $(\delta\ c) = (1\ 0)$  or  $(2\ 1)$. However, as noted earlier,  $\delta\ge 3$.  Thus Case 2 is impossible. Since both cases are impossible, the equality  $\Pi_{\frac n4} = n$  never holds for any  $n\ge 20$.  END OF PROOF

The next challenge starts to show up, namely the question of the initial constants   $10\ \,20$   which appear in  Theorems 1  2.  How do we compute them? Another one is the relation between the prime divisors of  $n-d$  and the function  $\nu(n)$ which which appears in  $\Pi_{\nu(n)}$ -- e.g. we had  $\nu(n):=\frac n4$  in Theorem 1. Thus let's consider another simple stage to illustrate the issues more explicitely:

THEOREM 2   $\forall_{n\ge 56}\ \,\Pi_{\frac n8}\gt n$,   where  $n$  is understood to be an arbitrary natural number.

PROOF  The theorem holds for  $n=56$. Let  $n>56$,  and--by induction--we assume that the theorem holds for  $n-1$,  i.e.  $\Pi_{\frac{n-1}8}>n-1$.  Then.

$$\Pi_{\frac n8}\ge n$$

Thus all that we have to prove is that the equality is impossible. Indeed (arguing by a contradiction), assume that:

$$\Pi_{\frac n8} = n$$

Thus  $n$  is a product of  $2$  and of an odd integer (a product of odd primes), i.e.  $n\equiv 2\mod 4$.  Let's consider four cases:

Case 1:  assume that  $m:=n-2$ is such that  $8|m$  and  $m$  is not a power of  $2$,  i.e. there exists an odd prime divisor of  $m$.  Then  $p\le\frac m8<\frac n8$  hence  $p|n$.  Thus  $p|n-m=2$ which is wrong. Thus case 1 is impossible.

Case 2:  assume that  $n-2$  is a power of  $2$,  say  $n-2=2^d$ for an integer. Since  $n-2\ge 55$  it follows that  $d\ge 6$ and  $n\ge 66$.

REMARK 0  Cases 1 and 2 cover all possibilities when  $8|n-2$.

Let's go back to case 2. Let this time  $m:=n-10$.  Then  $8|m$  and  $m$  is not a power of  $2$.  To be precise,

$$m = 8\cdot(2^{d-3} - 1)$$

where  $2^{d-3}-1\ge 7$.  Let  $p$  be an arbitrary odd prime divisor of  $m$.  Then  $p\le\frac m8 < \frac n8$  hence  $p|n$.  Thus  $p|n-m = 10$.  It follows that  $p=5$. This means that

$$2^\delta - 1 = 5^c$$

where  $\delta = d-3\ge 3$  and  $c$  are positive integers. By simple equation 251, the only solution here is $(\delta\ c) = (1\ 0)$. That's a contradiction since  $\delta\ge 3\ne 1$.  Thus case 2 is impossible.

Case 3:  assume that  $m:=n-6$ is such that  $8|m$  and  $m$  is not of the form  $2^d\cdot3^c$,  i.e. there exists a prime divisor  $p>3$  of  $m$.  Then  $p\le\frac m8<\frac n8$  hence  $p|n$.  Thus  $p|n-m=6$,  a contradiction. Thus case 3 is impossible.

END OF PROOF