Attempted Translation of Kanold

An upper bound for Jacobsthal's Function

by Hans-Joachim Kanold

The untranslated article is here.

Paul Erdos:  Let $n$ be any integer.  Jacobsthal defines $g(n)$ to be the least integer so that amongst any $g(n)$ consecutive integers $a, a+1, \dots, a+g(n)-1$ there is at least one which is relatively prime to $n$

Paul Erdos has shown that for almost all $n$, the order of magnitude of $g(n)$ is the same as $\frac{n}{\varphi(n)}V(n)$ with  $\varphi(n)$, the well-known Euler function, and with $V(n)$  the number of different prime divisors of $n$.

"Almost all” means: all natural numbers $n$ with the exception of a set of natural density zero. Note [3] investigated relationships between $g(n)$ and prime numbers in arithmetic sequences.  (Note 3: KANOLD, H.-J.: Über Primzahlen in arithmetischen Folgen. Math. Ann. 156, 393-395 (1964), t57, 358-362 (t965).)

The present work aims to provide further estimates of $g(n)$ to be derived.

We can rely on square-free arguments $n>1$ restricted.  

Let:

(1) $n = p_1 p_2  \dots p_k$ with $2 \le p_1 < p_2 < \dots < p_k$ the canonical decomposition of $n$ into a product of prime numbers

Furthermore, for the shortcut:

(2)  $g(n) = g$ 

And: $g(p_1) = 2$ applies, we can now also assume $k > 1$.  Based on the definition of $g$ there exists (at least) one natural number $a$ so that:

(3)  $(a,n) = (a+g,n) = 1$;  and $(a + \gamma) > 1$ for $\gamma = 1, \dots, g-1$.

With the known function of Möbius we get a first basic relationship:

(4)  $g -1 = - \sum\limits_{d|n\text{, }d>1}\mu(d)\left(\left[\frac{a+g-1}{d}\right]-\left[\frac{a}{d}\right]\right)$

We now introduce the following abbreviated terms.

(5)  $\frac{a+g-1}{d} - \left[\frac{a+g-1}{d}\right] = \Theta_d; \frac{a}{d} - \left[\frac{a}{d}\right] = \vartheta_d$

From (3), for $d > 1$, it follows:

(6)  $0 \le \Theta_d \le 1 - \frac{2}{d}; \frac{1}{d} \le \vartheta_d \le 1 - \frac{1}{d}$ 

Because $\Theta_d = 1 - \frac{1}{d}$ leads to $d|a+g$, the second basic relationship is:

(7)  $(g-1)\frac{\varphi(n)}{n} = \sum_\limits{d|n\text{, }d>1}\mu(d)(\Theta_d - \vartheta_d)$

Let:

(8) $(a+\gamma, n) = \delta_{\gamma} \text{                      } (\gamma = 1, \dots, g-1)$

then according to (1) and (3) always:

(9)  $\delta_{\gamma} \ge p_1; \delta_{\gamma} | p_1 \dots p_k; V(\delta_{\gamma}) \ge 1$

We look at the product:

(10)  $\Delta = \delta_1 \delta_2 \dots \delta_{g-1}$

And ask: How often is $(1 \le K \le k)$  in $\Delta$? The number of multiples of $p_K$ which is not larger than $a + g - 1$ are given by $\left[\frac{a+g-1}{p_K}\right]$; the number of multiples that are not greater than $a$ is $\left[\frac{a}{p_K}\right]$.  So:

(11)  $\Delta = \prod\limits_{K=1}^k {p_K}^{\left[\frac{a+g-1}{p_K}\right] - \left[\frac{a}{p_K}\right]}$

The number of $\Delta$ contained prime numbers, each according to its multiplicity counted, is therefore 

(12)  $\sum\limits_{K=1}^{k} \left(\left[\frac{a+g-1}{p_K}\right] - \left[\frac{a}{p_K}\right]\right)$

According to (3) it is of course:

(13)  $\left[\frac{a}{p_K}\right] = \left[\frac{a-1}{p_K}\right]$; $\left[\frac{a+g-1}{p_K}\right] = \left[\frac{a+g}{p_K}\right]$

If we denote, as in the introduction, with $V(m)$ the number of different prime divisors of a natural number $m$, then we get the third basic relationship:

(14)  $\sum\limits_{\gamma = 1}^{g-1}V(\delta_{\gamma}) =$ $\sum\limits_{K=1}^{k}$ $\left(\left[\frac{a+g-1}{p_K}\right] - \left[\frac{a}{p_K}\right]\right)$  $= (g-1)\sum\limits_{K=1}^{k}\frac{1}{p_K} + \sum\limits_{K=1}^k \left( \vartheta_{p_K} - \Theta_{p_K} \right)$

For convenience, we will use:

(15)  $\frac{1}{g-1}\sum\limits_{\gamma=1}^{g-1} V(\delta_{\gamma})  -$ $\sum\limits_{K=1}^k \frac{1}{p_K} = \frac{1}{g-1}\sum\limits_{K=1}^k$ $( \vartheta_{p_K} - \Theta_{p_K} ) = \xi$

Let $d | n, d \nmid a + \gamma$ for $\gamma = 1, \dots, g-1$. It follows:

(16)  $\Theta_d - \vartheta_d = \frac{g-1}{d}$

Let $d | n, d> 1$ and:

(17)  $d | a +  \gamma_d$; $d | a + g - {\gamma`}_d$; $1 \le \gamma_d \le g - {\gamma'}_d$;  $1 \le \gamma_d, {\gamma'}_d \le \text{min}\{d-1, g-1\},$

It follows from (5):

(18)  $\vartheta_d = 1 - \frac{\gamma_d}{d}$; $\Theta_d = \frac{{\gamma'}_d - 1}{d}$; $\Theta_d - \vartheta_d = \frac{\gamma_d + {\gamma'}_d -1}{d} - 1$

According to (17):

(19)  $\gamma_d + {\gamma'}_d \le \text{min}\{g; 2d-2\}$;  $0 \le g - \gamma_d - {\gamma'}_d \equiv 0 \pmod d$

II. We now want to derive some simple estimates from the basic relationships.

Lemma 1:  It always applies that: $0 < |\xi| < 1$

Proof: We first show that $\xi(g-1)$ cannot be completely, therefore also in particular $\xi$ not equal to zero.  From (15) we get:

(20)  $p_1 \dots p_k \sum\limits_{\gamma = 1}^{g-1}V(\delta_{\gamma}) - (g-1) \sum\limits_{K=1}^k \frac{p_1\dots p_k}{p_K} = \xi(g-1)p_1 \dots p_k$ 

Furthermore,

(21)  $\left(\sum\limits_1^k \frac{p_1 \dots p_k}{p_K}, p_1 \dots p_k \right) = 1$

Would now be $\xi(g-1)$ completely, so would be:

(22)  $p_1 \dots p_k | g-1$; $g-1 \ge n$ 

But that would mean that there is a sequence of length $n$ which does not contain a number that is relatively prime to $n$, which is obviously a contradiction, since every such sequence has exactly $\varphi(n)$ to $n$ must contain coprime numbers.  It is always (cf. [2]).

(23)  $g \ge k+1$

After (6) and (15) follows

(24)  $|\xi| \le \frac{1}{k} \sum\limits_1^k |\vartheta_{p_K} - \Theta_{p_K}| \le \frac{1}{k} \sum\limits_1^k \left(1 -  \frac{1}{p_K} \right) < 1$

This completes the proof of the above proposition.

Lemma 2: Let $k>1$. For $\xi < 0$ applies $g < \frac{k-3}{|\xi|} -2$,  For $\xi > 0$ applies $g < \frac{k-1}{\xi} + 2$ 

Proof:  Because of $g(p_1 \dots p_k) > g \left(\frac{p_1 \dots p_k}{p_K} \right)$ for any $K \in \{ 1, \dots, k \}$  are for $d = p_K$ the prerequisites (17) are met.  According to (15), (17) to (19) we obtain:

(25)  $g - 1 = \frac{1}{\xi} \sum\limits_1^k \left( 1 - \frac{\gamma_{p_K} + {\gamma'}_{p_K} -1}{p_K} \right)$; $2 \le \gamma_{p_K} + {\gamma'}_{p_K} \equiv g \pmod {p_K}$

Let $\xi > 0$. If $\gamma_{p_K} + {\gamma'}_{p_K} = 2$ for all $K=1, \dots, k$ then for $k > 1$, $g - 2 > 0$

(26)  $n = p_1 \dots p_k | g-2$, $g-2 \ge n$

which leads to a contradiction as in (23). Thus, according to (9), (15) and (25) we have

(27)  $g-1 < \frac{1}{\xi} \left( k - \sum\limits_1^k \frac{1}{p_K} \right) = \frac{k}{\xi} - \frac{1}{\xi} \left( \frac{1}{g-1} \sum\limits_1^{g-1} V(\delta_{\gamma})  - \xi\right) \le \frac{k}{\xi} - \frac{1}{\xi}(1 - \xi);$ $g < \frac{k-1}{\xi} + 2$ 

Let $\xi < 0$. From (25) we obtain (28):

(28) $g-1 = \frac{1}{|\xi|} \left( \sum\limits_1^k \frac{\gamma_{p_K} - {\gamma'}_{p_K}-1}{p_K} -k \right)$

Assume $\gamma_{p_K} + {\gamma'}_{p_K} = 2p_K - 2$ [cf. (19)] for $K=1, \dots, k$, so would be

(29)  $p_K | g - 2p_K+2$; $n=p_1 \dots p_k | g+2$; $g+2 \ge n$

Then there would be a sequence of natural numbers of length $n-3$, which does not contain a number that is coprime to $n$. We would have:

(30) $n - \varphi(n) \ge n - 3$; $3 \ge \varphi(n) = (p_1 - 1) \dots (p_k - 1)$

Because $p_3 -1 \ge 4$, it follows $k \le 2$; because of $|\xi| = \sum\limits_1^k \frac{1}{p_K} - \frac{1}{g-1}  \sum\limits_1^{g-1} V(\delta_{\gamma}) \le \sum\limits_1^k \frac{1}{p_K} - 1$ 

but delivers

(31)  $1 + |\xi| \le \sum\limits_1^k \frac{1}{p_K} \le \frac{1}{2} + \frac{1}{3}$

a contradiction. Therefore, according to (28)

(32) $g - 1 < \frac{1}{|\xi|} \left( \sum\limits_1^k \frac{2p_K - 3}{p_K} - k\right) = \frac{1}{|\xi|} \left( k - 3 \sum\limits_1^k \frac{1}{p_K} \right) \le \frac{1}{|\xi|} (k - 3 - 3|\xi|)$; $g < \frac{k-3}{|\xi|} - 2$

This proves the Lemma 2. A simple consequence is

Lemma 3: Let $k > 1$. For $\xi \ge \frac{1}{k+1}$ or $\xi \le \frac{-1}{k+3}$ follows $g \le k^2$

Proof: If $\xi \le \frac{-1}{k+3}$, then Lemma 2 provides:

(33)  $g < (k-3)(k+3) - 2 = k^2 - 11 < k^2$

If $\xi \ge \frac{1}{k+1}$, then follows:

(34)  $g < (k-1)(k+1) + 2 = k^2 + 1$; $g \le k^2$

From a slightly different structure is the:

Lemma 4: If $m$ and $n$ are coprime, then $g(mn) \le \text{min} \{ mg(n) - \varphi(m) + 1; ng(m) - \varphi(n) + 1  \}$

Proof: We imagine a sequence of length $mg(n) = mg$ given $a+1, a+2, \dots, a+mg.$ We arrange these $mg$ numbers in a rectangular scheme:

(35) 

$a+1 \qquad \qquad \qquad \qquad a+2, \dots, \dots, \dots, a+m$

$a+m+1 \qquad \qquad \qquad a+m+2, \dots,\dots, \dots, a+2m$

$\vdots$

$a+(g-1)m + 1 \qquad a+(g-1)m+2,\dots,\dots,\dots,a+gm$

Each column of (35) consists of numbers belonging to the same residue class $\text{mod}$ $m$. There are exactly $\varphi(m)$ columns whose numbers are all relatively prime to $m$ Let the residue classes be:

(36) $a + r_1, a + r_2, \dots, a + r_{\varphi}$ $(\varphi = \varphi(m))$ 

with $1 \le r_1 < r_2 < \dots < r_{\varphi} \le m$ is relatively prime to $m$. We then extract the following sub-scheme from (35).

(37)

$a + r_1\qquad \qquad \qquad \qquad a + r_2, \dots, \dots, \dots, a + r_{\varphi}$

$a + m + r_1\qquad \qquad \qquad a + m + r_2, \dots, \dots, \dots, a + m + r_{\varphi}$

$\vdots$

$a+(g-1)m + r_1\qquad a+(g-1)m + r_2, \dots, \dots, \dots, a+(g-1)m + r_{\varphi}$

All numbers in (37) are relatively prime to $m$. Now consider any column of (37).

(38)  $a + \gamma m + r_i$; $\gamma = 0, 1, \dots, g-1$; $i \in \{ 1, \dots, \varphi \}$

Because of $(m,n)=1$, there exists a natural number $v_i$  such that:

(39) $a + r_i + v_in \equiv 0 \pmod m$

is fulfilled. Then we look at the consecutive numbers:

(40)  $\frac{a + r_i + v_i n}{m} + \gamma$

Among these numbers, there is always at least one that is relatively prime to $n$. Therefore, in each column of (37) there is at least one number that is relatively prime to $mn$. This gives

(41)  $g(mn) \le mg(n) - \varphi(m) + 1$

Because of the symmetry in $m$ and $n$, Lemma 4 is proved. If $m$ is equal to a prime number $p$, then we get

(42)  $g(pn) \le pg(n) - p + 2$

III We now want to estimate the function $p$ for small values ​​of $k$. We start with

Lemma 5: If $(n,6)=1$; $1 < k \le 11$; Then $g < 2k$

Proof: According to the prerequisite $p_1 \ge 5$. For $k=2,3,4$ is from the numbers $a + 1, a + 2, \dots, a + k + 1$ at most one by $p_1,$ or $p_2, \dots,$ or $p_k$ divisible.

So we have for $1 < k \le 4$

(43)  $k+1 \le g \le k+1$, i.e. $g = k+1 < 2k$

Assume: $5 \le k \le 11$. We have $p_3 \ge 11$ ; $p_4 \ge 13$; $p_5 \ge 17$; $\dots$ so

(44)  $p_{k-2} > 2k$ for $k = 5, \dots 11$

Now consider a sequence of length $2k-1$:

(45)  $a+1, a+2, \dots, a + 2k-1$,

then at most one of these numbers is divisible by $p_k$ or $p_{k-1}$ or $p_{k-2}$.  If $g-1 \ge 2k - 1$, then we could choose the sequence (45) such that of these numbers at least $2k-4$ are not relatively prime to $p_1 p_2 \dots p_{k-3}$. Since none of the prime numbers $p_3 \dots p_{k-3}$ can be contained in more than two of the numbers (45), at least 6 of these numbers must not be relatively prime to $p_1 p_2$. For $k\le 7$ at most two of the numbers are divided by $p_2$ divisible; then follows $2k-1 \ge 16$, thus a contradiction to $k \le 7$. For $k=8, 9, 10, 11$ applies instead of (44)

(46) $p_{k-3} > 2k$

Of the numbers (45) at least $2k-5$ are not relatively prime to $p_1 \dots p_{k-4}$ At least 7 of these numbers do not have to be relatively prime to $p_1 p_2$, at least 4 must be divisible by $p_1$. This gives $p_1 = 5$; $k \ge 9$. For $k=9, 10, 11$:

(47) $p_{k-4} \ge 2k-1$

Of the numbers (45) at least $2k-6$ are not relatively prime to $p_1 \dots p_{k-5}$ At least 8 are not coprime to $p_1 p_2$, at least 5 must be divisible by  $p_1 = 5$. Only the following case remains:

(48) $k=11$; $a+1 \equiv 0 \pmod 5$

Assume that $g(n) \ge 2k-1$ so that none of the numbers $a + \gamma + v\times 5$ $(\gamma = 2, 3, 4, 5$; $v = 0, 1, 2, 3)$ is coprime to $p_2 \dots p_{11}$ At most, 3 of these numbers are divisible by $p_2 \ge 7$ since the numbers are in the range of  $a+2+0*5=a+2$ to $a+5+3*5 = a+20$ and if more than $3$ were divisible by $p_2$, then the fourth  such integer would be greater than $a+2+3*7 = a+23$ which is impossible since it is outside of the range; of the remaining $13$ numbers, one is divisible by $p_{11}$ or $p_{10}$ or $p_9, \dots$ or $p_6 (\ge 19)$ So at least $7$ numbers must be divisibly by $p_3, p_4, p_5$. But $p_3 \ge 11$ which is a contradiction since at most $3$ can be divided by $p_5 > p_4 > p_3 \ge 11$. This completes the proof for Lemma 5.

Theorem 1: For  $1 < k \le 12$, it follows that $g\le k^2$

Proof: Assume $(n,6)=1$ According to Lemma 5, for $1 < k \le 11$

(49)  $g \le 2k - 1 < k^2$

For $k=12, p_1 \ge 7$ we receive directly

(50) $\sum\limits_1^{12} \frac{1}{p_K} < 1 - \frac{1}{3}$

which according to Lemma 2 leads to

(51)  $g \le 34 < k^2$

The case $k=12, p_1 = 5$ leads according to lemmas 4 and 5 to 

(52)  $g \le 5(2\times 11 - 1)  - 3 = 102 < k^2$

Assume $(n,6)=2$. Lemma 4 leads to 

(53)  $g \le 2g(p_2 \dots p_k)$ 

so for $k=2$, $g \le 4 = k^2$ and for $3 \le k \le 12$ according to (49) to

(54)  $g \le 2(2k-3) = 4k-6 < k^2$

Let $(n,6)=3$. One can immediately see that for $k=2$ the equation:

(55)  $g=3$ $(<k^2)$

applies; for $3 \le k \le 12$ is according to (49) and Lemma 4

(56)  $g \le 3(2k-3)-1 = 6k - 10 < k^2$

Finally, $(n,6)=6$, i.e. $p_1 = 2, p_2 = 3$ According to (53) $k=2$ the equation $g=2^2$ fulfilled; for $k=3$ is according to (55) immediately $g \le 2\times 3 < 3^2$ is evident. For $4 \le k \le 13$ we obtain from (56), with $k-1$ instead of $k$,

(57)  $g \le 12k -32 \le k^2$ for $k = 4, 8, 9, 10, 11, 12, 13$

We therefore still have to prove the theorem in the cases

(58)  $k = 5,6,7$; $p_1 = 2, p_2 = 3$

where according to (53) it is sufficient to show that

(59)  $g(p_3 \dots p_k) \le \frac{k^2}{2}$

is fulfilled. According to the fourth proposition,

(60)  $g(p_3 \dots p_k) \le 3g(p_3 \dots p_k) - 1$ 

There $p_3 \ge 5$ must be is for $k=5,6$ also $g(p_3 \dots p_k) = k-1$ what to

(61)  $g \le 2 \times 3(k-1) - 2 = 6k - 8 < k^2$

Finally, for k = 7 we have the estimates

(62)  $g(p_3 \dots p_7) \le 7 = k$; $g \le 6\times 7 - 2 = 40 < k^2$

This proves Theorem 1.

IV. In this section we want to obtain more general results. We start with

Theorem 2: Let $q_1, q_2, \dots$ be the $i$th prime numbers ordered by size; $n=p_1 \dots p_k$ with $p_1 < \dots < p_k$ given such that $p_1 \ge q_{l +  1}, 1 < k \le l^e$ is valid. Then $g \le k^2$

Proof: From B. Rosser [4]:

(Note: B. Rosser showed that the $n$th prime is greater than $n \log n$: link, Pierre Dusart improved on this result in 1999: link)

(63)  $\sum\limits_{v=1}^m \frac{1}{q_v} = \log \log m + B_m$, $B_m > B_{m+1}$ $(m \ge 2)$;

for $m=1$ of course applies 

(64) $\sum\limits_{v=1}^1 \frac{1}{q_v} = \frac{1}{2}$.

For $l \ge 1$ we get, with $q_{l+1} = p_1 < \dots < p_k$,

(65) $\sum\limits_1^k \frac{1}{p_K} \le \sum\limits_{v=l+1}^{l+k} \frac{1}{q_v} = \log \frac{\log( l + k)}{\log l} - \sum\limits_{\gamma = l}^{l+k-1} (B_{\gamma}  - B_{\gamma + 1}),$

According to (63)

(66)  $B_{\gamma} - B_{\gamma + 1} = \log \log (\gamma + 1)  - \log \log \gamma  - \frac{1}{q_{\gamma + 1}}$.

According to Taylor’s theorem,

(67)  $\log \log(\gamma + 1) = \log \log \gamma + \frac{1}{\gamma \log \gamma} - \frac{1}{2} \frac{1 + \log(\gamma + \vartheta_{\gamma})}{(\gamma + \vartheta_{\gamma})^2 {\log}^2 (\gamma + \vartheta_{\gamma}) }$ 

with a certain $\vartheta_{\gamma}$ where $0 < \vartheta_{\gamma} < 1$. This means

(68) $\sum\limits_1^k \frac{1}{p_K} \le \log \frac{\log(l + k)}{\log l} - \sum\limits_{\gamma = l}^{l+k-1} \left( \frac{1}{\gamma \log \gamma} - \frac{1}{q_{\gamma + 1}} - \frac{1}{2} \frac{1 + \log \gamma}{{\gamma}^2 {\log}^2 \gamma} \right)$ 

Note that according to B. Rosser [4]

(69)  $q_{\gamma + 1} > (\gamma + 1)\log(\gamma + 1)$

is fulfilled, then (68) becomes

(70) $\sum\limits_1^k \frac{1}{p_K} < \log \frac{\log(l+k)}{\log l} - \frac{1}{l \log l} + \frac{1}{(l + k)\log(l+k)} + \frac{1}{2} \sum\limits_l^{l+k-1} \frac{1 + \log \gamma}{{\gamma}^2 {\log}^2 \gamma}$,

because still

(71) $\sum\limits_{\gamma = 1}^{l+k-1} \frac{1 + \log \gamma}{{\gamma}^2 {\log}^2 l} \le \frac{1 + \log l}{l^2 {\log}^2 l} + \int_l^{l+k} \frac{1 + \log x}{x^2 {\log}^2 x}{d}x = \frac{1}{l \log l} \left(1 + \frac{1}{l} + \frac{1}{l \log l} + \frac{1}{l \log l} \right) - \frac{1}{(l+k)\log(l + k)}$ 

is correct, (70) also provides the estimate

(72) $\sum\limits_{1}^{k} \frac{1}{p_K} < \log \frac{\log (l + k)}{\log l} + \frac{1}{2(l+k)\log(l+k)} - \frac{1}{2l\log l}\left(1 - \frac{1}{l} - \frac{1}{l \log l}\right) < \log\left( \frac{\log k}{\log l} + \frac{l}{k \log l} \right) + \frac{1}{2(l+k)\log(l + k)} - \frac{1}{2l\log l}\left( 1 - \frac{1}{l} - \frac{1}{l \log l} \right)$.

From Theorem 1, we can assume that $k \ge 13$; $l \ge 3$.

Let:

(73)  $13 \le k \le l^{7/3} - l$, so $l \ge 4$

then we have

(74)  $\frac{\log(l+k)}{\log l} \le 2.334$; $\sum\limits_1^k \frac{1}{p_K} < \log 2.334 + \frac{1}{2(l+k)\log l} - \frac{1}{2l \log l}\left( 1 - \frac{1}{l} - \frac{1}{l \log l}  \right) < 0.85 + \frac{1}{4^2 \log 4} < 0.9 = 1 - \frac{1}{10}$,

which according to Proposition 2 leads to $g < 10(k-1) +2 < 10k < k^2$. 

Let: 

(75)  $l^{7/3} - l < k \le l^{7/3}$, so $l \ge 4$;

According to (72) it follows that $\log \frac{\log(l + l^{7/3})}{\log l} < \log \left( \frac{7}{3} + \frac{1}{l^{4/3}\log l} \right) < \log \frac{7}{3} + \frac{3}{7l^{4/3}\log l}$ also:

(76) $\sum\limits_1^k \frac{1}{p_K} < \log \frac{7}{3} + \frac{3}{7l^{4/3}\log l} + \frac{3}{7l^{7/3}\times 2\log l} - \frac{1}{2l\log l} \left( 1 - \frac{1}{l} - \frac{1}{l \log l} - \frac{6}{7l^{1/3}} - \frac{3}{7l^{4/3}} \right) < 0.9$,

and again to $g < 10k < k^2$ so that: 

(77) $l^{7/3} < k \le l^{5/2} -l$ and $\log \frac{\log(l+k)}{\log l} \le \log \frac{5}{2} < 0.92$

In this case we see from (72)

(78)  $\sum\limits_1^k \frac{1}{p_K} < 0.92 + \frac{3}{14l^{7/3}\log l} - \frac{1}{2l\log l} \left( 1 - \frac{1}{l} - \frac{1}{l \log l}  \right) = 0.92 - \frac{1}{2l \log l} \left( 1 - \frac{1}{l} - \frac{1}{l \log l} - \frac{3}{7l^{4/3}} \right) < 0.92 - \frac{7}{60l \log l} < 1 - \frac{1}{12.5}$

which according to Proposition 2 immediately

(79)  $g < 12.5k < k^2$

delivers. For

(80)  $\text{max} \{ l^{7/3}; l^{5/2}-1  \} < k < l^{5/2}$

becomes

(81)  $\sum \frac{1}{p_K} < 0.92 + \frac{2}{5l^{3/2}\log l} + \frac{1}{5l^{5/2}\log l} - \frac{1}{2l\log l} \left( 1 - \frac{1}{l} - \frac{1}{l \log l} \right) = 0.92 - \frac{1}{2l\log l} \left( 1 - \frac{1}{l} - \frac{1}{l \log l} - \frac{2}{5l^{3/2}} - \frac{4}{5l^{1/2}} \right) < 0.92 + \frac{0.1809}{2l \log l} < 0.95$

for $k \ge 19$ according to Lemma 3, the inequality $g \le k^2$ fulfilled. For $k \le 18$ is $l^{7/3}< 18$ only fulfilled for $l \le 3$  

Then it follows from (80)

(82)  $12 < \text{max} \{ 9 \sqrt[3]{3}; 9\sqrt{3} - 3 \} < k < 9\sqrt{13} < 16$

In the cases $l=3, k=13, 14, 15$ we can estimate directly: 

(83)  $g < 5k < k^2$

We can assume for the rest of the proof

(84)  $l^{5/2} < k \le l^e$

According to (72)

(85)  $\sum \frac{1}{p_K}$ 

$< \log \left( e + \frac{1}{ll^{e-1}\log l}  \right) + \frac{1}{5l^{5/2}\log l} - \frac{1}{2l\log l} \left( 1 - \frac{1}{l} - \frac{1}{l \log l} \right)$  $< 1 - \frac{1}{2l \log l}$ $\left( 1 - \frac{1}{l} - \frac{1}{l \log l} - \frac{2}{el^{e-2}} - \frac{2}{5l^{3/2}} \right)$

For $l \ge 5$ is $\frac{1}{l} + \frac{1}{l \log l} + \frac{2}{el^{e-2}} + \frac{2}{5l^{3/2}} < \frac{1}{5}  + \frac{1}{8} + \frac{2}{2.7 \times 5^{3/2}} + \frac{2}{25 \times \sqrt{5}} < 0.62$ i.e. we have the estimates

(86)  $\sum \frac{1}{p_K} < 1 - \frac{0.38}{2l \log l}$; $g < \frac{l \log l}{0.19} k \le k^2$

because $\frac{\log l}{l} \le \frac{\log 5}{5} < 0.322 < 0.19l^{1/2}$ is fulfilled. For $l = 3,4$ we get from (84) and $p_1 \ge q_{l + 1}$

(87)  $p_1 = 7$ or $11$; $16 \le k \le 19$ or $33 \le k \le 53$.

For $l=3$, it follows:

(88)  $\sum\limits_1^{19} \frac{1}{p_K} \le \frac{1}{7} + \frac{1}{11} + \dots + \frac{1}{79} < 0.8$; $g < 5k < k^2$

For $l=4$, it follows:

(89) $\sum\limits_1^{53} \frac{1}{p_K} \le \frac{1}{11} + \frac{1}{13} + \dots + \frac{1}{269} < 1 - \frac{1}{30}$; $g < 30k < k^2$

This confirms the second theorem.

Theorem 3:  Let $n = p_1 \dots p_k$, $1 < k^{0.6} < p_1 < \dots < p_k$ then $g(p_1 \dots p_k) \le k^2$.

Proof: We write 

(90) $\pi(k^{0.6}) = l$

and know from Theorem 2 that $k \le l^{e}$ the assertion of the above theorem follows.  Therefore we now ask the question: For which $l$ is:

(91)  $l < \pi(l^{0.6e})$

fulfilled? Assume first, with $0.6e = \alpha (1.63 < \alpha < 1.632)$

(92)  $l^{\alpha} \ge e^{100}$

Then, according to B. Rosser [4]:

(93)  $\pi(l^{\alpha}) \ge \frac{l^{\alpha}}{\alpha \log l + 2}$

We get

(94)  $l < \frac{l^{\alpha}}{\alpha \log l + 2} \le \pi(l^{\alpha})$

Because $\alpha l \log l + 2l < l^{\alpha}$, $\frac{\alpha}{a-1} \frac{\log l^{\alpha-1}}{l^{\alpha-1}} + \frac{2}{l^{\alpha - 1}} \le \frac{1.632}{0.63} \times \frac{10}{e^{10}} + \frac{2}{e^{10}} < 1$, it is fulfilled. Assume:

(95)  $17 \le l^{\alpha} < e^{100}$

Then we have, according to B. ROSSER [4]

(96)  $\pi(l^{\alpha}) \ge \frac{l^{\alpha}}{\alpha \log l} > l$; $\alpha \log l < l^{\alpha -1}$

because the function $\frac{\log x}{x}$ for $1 < x$ is positive and at $x = e$ the maximum $\frac{1}{e}$ 

Because the above adjustment is equivalent to:

(97)  $\frac{\alpha}{\alpha - 1} \times \frac{\log l^{\alpha - 1}}{l^{\alpha - 1}} \le \frac{\alpha}{\alpha - 1} \times \frac{1}{e} \le \frac{0.6e}{0.63} \times \frac{1}{e} = \frac{60}{63} < 1$

Is now

(98)  $l^{1.63} < l^{\alpha} < 17$; $l < 6$

then we investigate for which $l \le 5$ according to (91)

(99)  $l < \pi(l^{1.6})$

applies. It is

(100)  $5^{1.6} > 13$; $\pi(13) = 6$

For $l=2,3,4$ we carry out the proof directly with estimation of $\sum\limits_1^{k} \frac{1}{p_K}$ and application of Proposition 2. Because of Theorem 1 we can $k \ge 13$; $l = \pi(k^{0.6}) \ge \pi(\sqrt{13}) \ge \pi(3) = 2$ This proves Theorem 3.

Theorem 4: For all $n = p_1 \dots p_k$, $g(n) \le 2^k = \sum\limits_{d|n} 1$. For $k \ge e^{50}$ is $g(n) < 2^{\sqrt{k}}$.

Proof: For $k=1,2$ the statement is correct. So let $k \ge 3$. Assume that:

(101) $p_1 < \dots < p_i \le (k-i)^{0.6} < p_{i+1} < \dots < p_k$.

From Lemma 4, it follows that:

(102)  $g(n) \le p_1 \dots p_i  g(p_{i+1} \dots p_k) - (p_1 - 1) \dots (p_i - 1) + 1$

From Theorem 3 we get

(103)  $g(n) \le p_1 \dots p_i(k-i)^2$

It is known

(104)  $p_1 \dots p_i < 4^{(k-i)^{0.6}}$

Therefore follows

(105)  $g(n) < 2^{2(k-i)^{0.6}} \times (k-i)^2 \le 2^k$ 

if

(106)  $2^{2(k-i)^{0.6}} \times (k-i)^2 \le 2^{2k^{0.6}} \times k^2 \le 2^k$; $k^2 \le 2^{k^{0.6}(k^{0.4}-2)}$; $2\log k \le k^{0.6} (k^{0.4} - 2) \log 2$; $\frac{\log k^{0.6}}{k^{0.6}}\le 0.3 \times (k^{0.4} - 2) \log 2$

applies. Because $\frac{\log k^{0.6}}{k^{0.6}} \le \frac{1}{e} < 0.368 < 0.3 \times 1.8 \log 2$ is correct, it follows that $k^{0.4} - 2 \ge 1.8$; $k \ge 29$ immediately the claim. For $3 \le k \le 12$ the assertion of the theorem is also fulfilled, namely for $k=4, \dots, 12$ according to Theorem 1, for $k=3$ because of $g(p_1 p_2 p_3) \le 6$. Now $13 \le k \le 28$ provided. We get according to (101)

(107)  $p_i \le 28^{0.6} < 10$

With this we have

(108)  $p_1 \dots p_i \le 2 \times 3 \times 5 \times 7 = 210$; $g(n) < 210k^2$

For $k \ge 18$ is:

(109)  $210k^2 \le 2^k$

Because $\frac{\log 210}{k} + \frac{2 \log k}{k} < \frac{5.35 + 5.8}{18} < \log 2$ is fulfilled. For $13 \le k \le 17$, it follows:

(110) $p_i \le 17^{0.6} < 7$; $p_1 \dots p_i \le 30$; $g(n) < 30k^2 \le 2^k$,

because $\frac{\log 30 + 2 \log k}{k} < \frac{3.41 + 5.14}{13} > \log 2$ applies:

For $k \ge e^{50}$ we write, similar to the proof of Theorem 3:

(111)  $l = \pi(k^{0.48})$; $l < \pi(l^{0.48e})$; $\alpha = 0.48e$ $(1.3 < \alpha < 1.31)$

According to B. Rosser [4]

(112)  $\pi(l^{\alpha}) \ge \frac{l^{\alpha}}{\alpha \log l + 2}$; $\frac{\alpha}{\alpha - 1} \times \frac{\log l^{\alpha - 1}}{l^{\alpha - 1}} + \frac{2}{l^{\alpha - 1}} <  \frac{1.31}{0.3} \times \frac{\log l^{\alpha - 1}}{l^{\alpha - 1}} + \frac{2}{l^{\alpha - 1}}$; $k^{0.48} \ge e^{24}$; $l \ge \pi(e^{24}) \ge \frac{e^{24}}{26}$; $l^{\alpha - 1} > l^{0.3} > \left( \frac{e^{24}}{26} \right)^{0.3} > e^{20\times 0.3} = e^6 > 400$; $\frac{131}{30} \times \frac{6}{400} + \frac{2}{400} < 1$

We therefore have the following result:

(113)  For $k \ge e^{50}$; $k^{0.48}< p_1 < \dots < p_k$, it follows that $g(p_1 \dots p_k) \le k^2$

Let $p_1 < \dots < p_j \le (k-j)^{0.48} < p_{j+1} < \dots < p_k$ so it follows that:

$g \le p_1 \dots p_j \times (k-j)^2 < 2k^{0.48} \times k^2 \le 2^{\sqrt{k}}$ 

Then: $2k^{0.48} \log 2 + 2\log k \le \sqrt{k}\log 2$; $2\log k \le k^{0.48}(k^{0.02}-2) \log 2$; $k^{0.02} \ge e > 2.7$; $2 \log k \le 0.7 \times k^{0.48} \log 2$; $\frac{1}{0.24} \times \frac{\log k^{0.48}}{k^{0.48}} < 0.7 \log 2$; $\frac{1}{0.24} \times \frac{24}{e^{24}} < 0.7 \log 2$ is valid. This proves Theorem 4.

As a simple consequence of the last theorem we get:

(114) For $k \ge 18$, it follows that $n > g^4$

Proof: 

Since $k \ge 18$:

(115)  $n \ge 2 \times 3 \times \dots \times 31 \times 37^{k - 11} > 2^{42}\times 37^{k-11} > 2^{5k-18} \ge 2^{4k} \ge g^4$

If we imagine the divisors $d$ of $n$ arranged according to size, then $2^{k-1}$ this divisor is greater than $\sqrt{n} > g^2$ if $k \ge 18$ is fulfilled. For each divisor $d$ from $n$ with $d > g^2$ is either according to (16):

(116) $0 < \Theta_d - \vartheta_d = \frac{g-1}{d} < \frac{g-1}{g^2} < \frac{1}{g}$

or according to (17) and (18)

(117)  $\Theta_d - \vartheta_d = \frac{g-1}{d} - 1$; $0 < 1 + \Theta_d - \vartheta_d < \frac{g-1}{g^2} < \frac{1}{g}$

The estimates in this work can be improved with greater computational effort, in particular it can be shown that $1 < k^{1/2} < p_1 < \dots < p_k$ also $g < k^{1.7}$ it follows for $k > 659^2$ This bound for k can probably be lowered by further calculations.

 

 

 

Solution to the Depressed Cubic Equation

Theorem 9:  Depressed Cubic Equation

If $x^3 + bx = c$, then:

$$x = \sqrt[3]{\frac{c}{2} + \sqrt{\dfrac{c^2}{4} + \frac{b^3}{27}}} + \sqrt[3]{\frac{c}{2} - \sqrt{\dfrac{c^2}{4} + \frac{b^3}{27}}}$$

Proof:

(1)  $x^3 = -bx + c$

(2)  There exists real numbers $u,v$ such that $x = u + v$

(3)  $(u+v)^3 = u^3 + 3u^2v + 3uv^2 + v^3 = 3uv(u+v) + (u^3 + v^3)$

(4)  Since $b = -3uv$, it follows that $v = \dfrac{-b}{3u}$

(5)  $c = u^3 + v^3 = u^3 + \dfrac{-b^3}{27u^3}$

(6)  It follows that:

$$u^6  - u^3c  + \dfrac{-b^3}{27}= 0$$

(7)  Using the solution for the quadratic equation, we have:

$$u^3 = \frac{c \pm \sqrt{c^2 + \frac{4b^3}{27}}}{2} = \frac{c}{2} \pm \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

(8)  From the symmetry of the equation, we can assume that:

$$u^3 = \frac{c}{2} + \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

$$v^3 = c - u^3 = c - \frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3} = \frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

(9)  $x = u + v = \sqrt[3]{\frac{c}{2} + \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}} + \sqrt[3]{\frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}}$

Solution to the Quadratic Equation

Theorem 8:  If $ax^2 + bx + c=0$, then:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Proof:

(1)  $x^2 + \dfrac{bx}{a} + \dfrac{c}{a} = 0$

(2)  $x^2 + \dfrac{bx}{a} = -\dfrac{c}{a}$

(3)  $x^2 + \dfrac{bx}{a} + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$

(4)  $\left(x + \dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$

(5)  $x + \dfrac{b}{2a} = \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2}$

(6) $x = -\dfrac{b}{2a} \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2} = \dfrac{-b \pm  \sqrt{b^2 - 4ac}}{2a}$

The Square Root of 2 is irrational

Theorem 7: The Square Root of $2$ cannot be represented by the ratio of two integers

Proof:

(1)  Assume that the squre root of $2$ could be represented by a ratio of two integers with $a$ and $b$ the reduced form.

$$\frac{a}{b} = \sqrt{2}$$  

(2)  Squaring both sides:

$$a^2 = 2b^2$$

(3)  Since $a$ must be even, there exists $c$ such that $a=2c$ and:

$$a^2 = (2c)^2 = 4c^2 = 2b^2$$

(4)  It follows that $b$ must be even since:

$$2c^2 = b^2$$

(5)  Then there exists $d$ such that $b=2d$ and:

$$\sqrt{2} = \frac{2c}{2d} = \frac{c}{d}$$

(6)  But then we have a contradiction since $a$ and $b$ can be reduced to $c$ and $d$.

Infinitude of Prime Numbers

Theorem 6:  Infinitude of Prime Numbers

There are an infinite number of prime numbers.

Proof:

(1)  Assume that there is a a finite number of primes. 

(2)  Let $P$ be the product of all primes.

(3)  Let $q = P+1$

(4)  By the Fundamental Theorem of Arithmetic, either $q$ is prime or $q$ is a product of primes.

(5)  If $q$ is prime, then $q | (P + 1) - P = 1$ which s impossible.

(6)  If $r$ is a prime that divides $q$, then we have the same problem since $r | (P + 1) - P = 1$

Fundamental Theorem of Arithmetic

Theorem 5:  The Fundamental Theorem of Arithmetic

Every integer greater than $1$ is either prime or can be represented as the unique product of powers of prime numbers.

Proof:

(1)  First, prove existence.  It is true for the base case where $n=2$ since $2$ is a prime.

(2)  Assume that it is true for all $i$ with $i \le n$

(3)  For $n+1$, if $n+1$ is prime, then it is true.  If $n+1$ is not prime, then there exists integers $a,b$ such that $n+1 = ab$ with $a \le n$ and $b \le n$.  Since $a \le n$ and $b \le n$, they can be represented by a prime or a product of primes with $a=p_1p_2\dots{p_n}$ and $b=q_1q_2\dots{q_n}$

(4)  Then $n+1 = ab = p_1p_2\dots{p_n}q_1q_2\dots{q_n}$ which means that $n+1$ can be represented as a product of powers of prime numbers.

(5)  Assume that there is an integer that has two different product of powers of prime numbers.

(6)  Let $y$ be the least integer where this is true so that $y = p_1p_2\dots{p_m} = q_1q_2\dots{q_n}$ with $p_1\dots{p_m}$  and $q_1\dots{q_n}$.

(7)  From Euclid's Generalized Lemma, $p_1$ divides $q_i$ where $1 \le i \le n$.  But, then $\frac{y}{p_1} = p_2\dots{p_m} = \frac{q_1q_2\dots{q_n}}{p_1}$ which is smaller than $n$ and contradicts our assumption in step(6).

Euclid's Lemma

Theorem 4:  Euclid's Lemma

If a prime $p$ divides the product $ab$ of two integers $a$ and $b$, then $p$ divides $a$ or $p$ divides $b$

Proof:

(1)  Assume that $p | ab$ but $p \nmid a$

(2)  From Bezout's Identity, there exists $r,s$ such that $rp + sa = 1$

(3)  Multiply both sides by $b$ to get:  $rpb + sab = b$

(4)  Since $p | ab$, it follows that $p | b$

Corollary 4.1:  Euclid's Generalized Lemma

If a prime $p$ divides $a_1a_2\dots{a_m}$, then $p$ divides $a_i$ where $1 \le i \le m$

Proof:

(1)  Base Case:  The base case is Euclid's Lemma.  If $p | a_1a_2$, then $p | a_1$ or $p | a_2$

(2)  Assume that up to $n$ if $p | a_1a_2\dots{a_n}$, then $p|a_i$ with $1 \le i \le n$

(3)  Let $b = a_1a_2\dots{a_n}$.  

(4)  Inductive Case:  From Euclid's Lemma, if $p | ba_{n+1}$, then $p | b$ or $p|a_{n+1}$.