**2. Proof of the theorem 1**

In this section, we prove theorem 1. Components of the proof Elementary algebra basic operations such as subtraction from both sides and extraction of the common

*On the Analytical Properties of Prime Numbers DOI: http://dx.doi.org/10.5772/intechopen.109365*

factor with the binomial theorem form the foundations of the proof. Theorem 1 is an expression of a polynomial that shows the properties of numbers in the form *<sup>p</sup>* <sup>¼</sup> *ma<sup>m</sup>* <sup>þ</sup> *bm* <sup>þ</sup> 1,

*a*, *m*, *b*> 1∈ , so it can be used as a test to reveal the prime number in the form *mam* <sup>þ</sup> *bm* <sup>þ</sup> 1. In addition to that, it is used to prove the results in the next section where it plays an essential role in the proofs.

**THEOREM 1.** if M is a prime where <sup>M</sup> <sup>¼</sup> *ma<sup>m</sup>* <sup>þ</sup> *bm* <sup>þ</sup> <sup>1</sup> *and a*, *<sup>m</sup>* <sup>&</sup>gt;1<sup>∈</sup> , *<sup>b</sup>* 6¼ 0∈ then

$$\begin{cases} \eta\_{(\boldsymbol{\lambda})} \equiv \chi\_{(\mathbf{x}, \boldsymbol{y})}(\boldsymbol{mod} \; \mathcal{M}) & \text{if } \; \mathcal{M} \; \text{and} \; \boldsymbol{\lambda} \; \text{is a prime where} \\\\ \eta\_{(\boldsymbol{\lambda})} \equiv \chi\_{(\boldsymbol{\lambda}; \mathbf{x}, \boldsymbol{y})}(\boldsymbol{mod} \; \mathcal{M}) & \text{if } \; \mathcal{M} \; \text{is a prime} \end{cases} \tag{2}$$

**Proof.** let <sup>M</sup> <sup>¼</sup> *ma<sup>m</sup>* <sup>þ</sup> *<sup>δ</sup> where m*, *<sup>a</sup>*, n>1<sup>∈</sup> and *<sup>δ</sup>* <sup>¼</sup> *bm* <sup>þ</sup> 1 where *<sup>b</sup>*∈ according to the binomial theorem, we find that

$$(\mathcal{M} + (-\delta))^n - m^n = (ma^m)^n - m^n = \sum\_{j=0}^{n-1} \binom{n}{j} \mathcal{M}^{n-j} (-\delta)^j$$

$$+ (-\delta)^n - m^n \tag{3}$$

Then

$$m^n(a^{mn} - 1) - ((-bm - 1)^n - m^n)$$

$$= \sum\_{j=1}^{n-1} \binom{n}{j} \mathcal{M}^{n-j}(-bm - 1)^j \tag{4}$$

Note that

$$\sum\_{j=0}^{n-1} \binom{n}{j} \mathcal{M}^{n-j} (-bm - 1)^j \equiv \mathbf{0} (mod \; \mathcal{M}) \tag{5}$$

Then from (4) and (5) we have that

$$a^{nm} - \mathbf{1} - ((-bm - \mathbf{1})^n - m^n) \equiv \mathbf{0} (mod \ \mathcal{M}) \tag{6}$$

Now we conclude that from Eq. (6)

$$a^{nm} \equiv \mathbf{1}(mod \; \mathcal{M}) \text{ if } \; and \; only \; \text{if } \; (-bm - 1)^n \equiv m^n(mod \; \mathcal{M}) \tag{7}$$

Suppose that *<sup>n</sup>* <sup>¼</sup> <sup>M</sup>�<sup>1</sup> *<sup>m</sup>* and M is a prime so

$$a^{\mathcal{M}-1} \equiv \mathbf{1}(mod \; \mathcal{M}) \; \text{if} \; \text{and} \; only \; \text{if} \; (-bm - 1)^{\frac{\mathcal{M}-1}{m}} \equiv m^{\frac{\mathcal{M}-1}{m}}(mod \; \mathcal{M}) \tag{8}$$

From the assumption M is a prime from Fermat's litter theorem see [Kenneth H. Rosen 2 pp. 161] we have that if <sup>M</sup> *is a prime* then *<sup>a</sup>*<sup>M</sup>�<sup>1</sup> � <sup>1</sup> ð Þ *mod*<sup>M</sup> where *<sup>a</sup>*>1. This means if M is a prime number, then from (8) we find that

$$(-bm - 1)^{\frac{\mathcal{M} - 1}{m}} \equiv m^{\frac{\mathcal{M} - 1}{m}} (mod \, \mathcal{M}) \tag{9}$$

$$(bm+1)^{\frac{\mathcal{M}-1}{m}} \equiv (-m)^{\frac{\mathcal{M}-1}{m}}(mod\ \mathcal{M})\tag{10}$$

$$(bm+1)^{\dot{\lambda}} - (-m)^{\dot{\lambda}} = \sum\_{j=0}^{\dot{\lambda}} \binom{\dot{\lambda}}{j} (bm)^{\dot{\lambda}-1} - (-m)^{\dot{\lambda}}$$

$$= (bm)^{\dot{\lambda}} - (-m)^{\dot{\lambda}} + \sum\_{j=1}^{\dot{\lambda}-1} \binom{\dot{\lambda}}{j} (bm)^{\dot{\lambda}-1} + 1 \tag{11}$$

$$(bm+\mathbf{1})^{\dot{\lambda}}-(-m)^{\dot{\lambda}}=(bm)^{\dot{\lambda}}-(-m)^{\dot{\lambda}}+\mathbf{1}$$

$$+\sum\_{j=1}^{i-1}\pi\_{j}b^{\lambda-j}m^{\lambda-j-1}(\mathcal{M}-\mathbf{1})$$

$$=(bm)^{\dot{\lambda}}-(-m)^{\dot{\lambda}}+\mathbf{1}+\left(\sum\_{j=1}^{i-1}\pi\_{j}b^{\lambda-j}m^{\lambda-j-1}\right)\mathcal{M}$$

$$-\sum\_{j}^{\dot{\lambda}-1}\pi\_{j}b^{\lambda-j}m^{\dot{\lambda}-j-1}\tag{12}$$

$$\left(\sum\_{j=1}^{\lambda-2} \pi\_j b^{\lambda-j} m^{\lambda-j-1}\right) \mathcal{M} \equiv \mathbf{0} (mod \; \mathcal{M}) \tag{13}$$

$$(bm)^{\lambda} - (-m)^{\lambda} - \sum\_{j=1}^{k-2} \pi\_j b^{\lambda - j} m^{\lambda - j - 1} - b + \mathbf{1} \equiv \mathbf{0} \\
(mod \; \mathcal{M}) \tag{14}$$

$$\sum\_{j=1}^{\lambda-2} \pi\_j b^{\lambda-j} m^{\lambda-j-1} \equiv (bm)^{\lambda} - (-m)^{\lambda} - b + \mathbf{1} (mod \; \mathcal{M}) \tag{15}$$

$$\text{Let be } \eta\_{(\vec{\boldsymbol{\alpha}})}(\mathbf{x}, \mathbf{y}) = \sum\_{j=1}^{\lambda - 2} \pi\_j \mathbf{x}^{\lambda - j} \mathbf{y}^{\lambda - j - 1} \text{ and } \boldsymbol{\chi}\_{(\mathbf{x}, \mathbf{y})} = (bm)^{\lambda} - (-m)^{\lambda} - b + \mathbf{1} \text{ So we}$$

$$\eta\_{(\lambda)}(\mathbf{x}, \mathbf{y}) \equiv \chi\_{(\mathbf{x}, \mathbf{y})}(\operatorname{mod } \mathcal{M}) \tag{16}$$

$$\lambda \left( (bm+\mathbf{1})^{\dot{\lambda}} - (-m)^{\dot{\lambda}} \right) = \lambda \left( (bm)^{\dot{\lambda}} - (-m)^{\dot{\lambda}} \right) + \lambda + \sum\_{j=1}^{\ell-1} \binom{\dot{\lambda}}{j} b^{\ell-j} m^{\dot{\lambda}-j-1} (\mathcal{M}-\mathbf{1})$$

$$= \lambda \left( (bm)^{\dot{\lambda}} - (-m)^{\dot{\lambda}} \right) + \lambda + \left( \sum\_{j=1}^{\ell-1} \binom{\dot{\lambda}}{j} b^{\ell-j} m^{\dot{\lambda}-j-1} \right) \mathcal{M}$$

$$- \sum\_{j=1}^{\ell-2} \binom{\dot{\lambda}}{j} b^{\dot{\lambda}-j} m^{\ell-j-1} - \lambda b \tag{17}$$

$$\left(\sum\_{j=1}^{\lambda-1} \binom{\lambda}{j} b^{\lambda-j} m^{\lambda-j-1} \right) \mathcal{M} \equiv (mod \; \mathcal{M}) \tag{18}$$

$$\lambda\left(\left(bm+\mathbf{1}\right)^{\lambda}-\left(-m\right)^{\lambda}\right)\equiv\mathbf{0}(mod\ \mathcal{M})\tag{19}$$

$$\sum\_{j=1}^{\lambda-2} \binom{\lambda}{j} b^{\lambda-j} m^{\lambda-j-1} \equiv \lambda (bm)^{\lambda} - \lambda (-m)^{\lambda} + \lambda - b\lambda (mod \mathcal{M}) \tag{20}$$

$$\eta\_{(\boldsymbol{\lambda})}(\boldsymbol{x}, \boldsymbol{y}) \equiv \chi\_{(\boldsymbol{\lambda}, \mathbf{x}, \boldsymbol{y})}(\boldsymbol{mod} \, \mathcal{M}) \tag{21}$$

$$\begin{cases} \eta\_{(\boldsymbol{\lambda})}(\mathbf{x}, \boldsymbol{\upmu}) \equiv \chi\_{(\boldsymbol{\upmu}, \mathbf{x})}(\boldsymbol{\upmu} \boldsymbol{\upmu} \boldsymbol{\upmu}) & \text{if } \boldsymbol{\upmu} \text{ and } \boldsymbol{\upmu} \text{ is a prime} \\\\ \eta\_{(\boldsymbol{\upmu})}(\mathbf{x}, \boldsymbol{\upmu}) \equiv \chi\_{(\boldsymbol{\upmu}, \mathbf{x})}(\boldsymbol{\upmu} \boldsymbol{\upmu} \boldsymbol{\upmu}) & \text{if } \boldsymbol{\upmu} \text{ is a prime} \end{cases} \tag{22}$$

**REMARK 1:** We note that the proof has little complexity, as we explicitly relied on the binomial theorem and elementary algebra to obtain Eq. (12). After that, Fermat's Litter Theorem was used, which is a theorem dating back to the year 1610.In 1610 Fermat wrote in a letter to Frenicle, that whenever p is prime p divides *<sup>a</sup>p*�<sup>1</sup> � 1 for all integers *a* not divisible p, a result now known as Fremat's little theorem, As equivalent formulation is the assertion that p divide *<sup>a</sup><sup>p</sup>* � *<sup>a</sup>* for all integers *<sup>a</sup>* , whenever p is prime. The question naturally arose as to whether the prime are the only integer exceeding that satisfy this criterion, but Carmichael pointed out in 1910 that 561 = 11 � 17 � 3 divides a *<sup>a</sup>*<sup>560</sup> � <sup>1</sup>ð Þ *mod* <sup>561</sup> now. A composite integer which satisfies *<sup>a</sup>n*�<sup>1</sup> � <sup>1</sup>ð Þ *mod n* for all positive integers *a* with g.c.d(a, n) = 1 is called a Carmichael number. For a related discussion see Kenneth H. Rose page (55). This means that Theorem 1 is not a definitive test, but it fails at the numbers Carmichael, but on the one hand we find that it is more general than those [2, 10] because of the variables m, a, b in the number *mam* <sup>þ</sup> *bm* <sup>þ</sup> 1. And we will explain this by proving results for Mersenne and Fermat numbers, which are special cases when the variable *b* takes a certain value.

**THEOREM 2.** *if <sup>λ</sup>* <sup>¼</sup> *am* þ �ð Þ<sup>1</sup> *<sup>σ</sup>* and *<sup>q</sup>* <sup>¼</sup> *mam* þ �ð Þ<sup>1</sup> *<sup>σ</sup> m* þ 1 where q is a prime and *a*, *m* >1∈ then

$$\begin{cases} \boldsymbol{\nu}\_{(m)} \equiv \boldsymbol{\lambda} - (-1)^{\sigma} \boldsymbol{\lambda}(mod \; q \;) \\ \qquad \text{if } \sigma = 1 \; and \; a > 1 \in \mathbb{N} \; and \; \text{if } \sigma = 2 \; then \; a \; is \; odd \\ \boldsymbol{\nu}\_{(m)} \equiv 2 \boldsymbol{\lambda} m^{\lambda} + \boldsymbol{\lambda} - (-1)^{\sigma} \boldsymbol{\lambda}(mod \; q \;) \; \text{if } \sigma = 2 \; and \; a \; is \; even \end{cases} \tag{23}$$

**Proof.** Let be *<sup>b</sup>* ¼ �ð Þ<sup>1</sup> *<sup>σ</sup>* From theorem 1 we have

$$\sum\_{j=1}^{\lambda-2} \binom{\lambda}{j} ((-1)^{\sigma})^{\lambda-j} m^{i-j-1} \equiv \lambda ((-1)^{\sigma} m)^{\lambda} - \lambda (-m)^{\lambda} + \lambda - (-1)^{\sigma} \lambda (\text{mod } \mathbf{q}) \tag{24}$$

Let be

$$\mathcal{W}\_{(m)} = \sum\_{j=1}^{\lambda-2} \binom{\lambda}{j} ((-1)^{\sigma})^{\lambda-j} m^{\lambda-j-1} \tag{25}$$

From Eq. (24) we get the following

$$\begin{cases} \begin{aligned} \boldsymbol{\Psi}\_{(m)} & \equiv \boldsymbol{\lambda} - \boldsymbol{\lambda}(-\mathbf{1})^{\sigma}(\boldsymbol{mod}\,\boldsymbol{q}) \\ \text{if } \sigma = \mathbf{1} \,\boldsymbol{and}\,\boldsymbol{a} > \mathbf{1} \in \mathbb{N} \,\boldsymbol{and} \,\,\boldsymbol{if}\,\,\sigma = \mathbf{2} \,\boldsymbol{a} \,\,\boldsymbol{is}\,\,\boldsymbol{odd} \\ \boldsymbol{\Psi}\_{(m)} & \equiv 2\boldsymbol{\lambda}\boldsymbol{m}^{\lambda} + \boldsymbol{\lambda} - \boldsymbol{\lambda}(-\mathbf{1})^{\sigma}(\boldsymbol{mod}\,\,\boldsymbol{q}) \\ \text{if } \sigma = \mathbf{2} \,\,\boldsymbol{and}\,\,\boldsymbol{a} \,\,\boldsymbol{is}\,\,\boldsymbol{even} \end{aligned} \tag{26}$$

**LEMMA 1**. let be *<sup>p</sup>* <sup>¼</sup> <sup>3</sup>*<sup>m</sup>* � 2 and <sup>M</sup> <sup>¼</sup> *<sup>m</sup>*3*<sup>m</sup>* � <sup>2</sup>*<sup>m</sup>* <sup>þ</sup> 1 where *<sup>p</sup>* and <sup>M</sup> is a prime number then

$$\sum\_{j=1}^{p-2} \pi\_j (-2)^{p-j} m^{p-j-1} \equiv \left( \mathfrak{Z}\_1^p \right) (-m)^p + \mathfrak{Z} (mod \; \mathcal{M}) \tag{27}$$

**Proof**. Let be in theorem 1 *a* ¼ 3 *and b* ¼ �2 *where p is a prime* then we get *λ* ¼ <sup>3</sup>*<sup>m</sup>* � <sup>2</sup> *and* <sup>M</sup> <sup>¼</sup> *<sup>m</sup>*3*<sup>m</sup>* � <sup>2</sup>*<sup>m</sup>* <sup>þ</sup> 1 and we have

*On the Analytical Properties of Prime Numbers DOI: http://dx.doi.org/10.5772/intechopen.109365*

$$\sum\_{j=1}^{p-2} \pi\_j (-2)^{p-j} m^{p-j-1} \equiv \left( 2^p\_1 \right) (-m)^p + \mathfrak{Z} (mod \; \mathcal{M}) \tag{28}$$

Then

$$\sum\_{j=1}^{M\_p-2} \pi\_j (-1)^{M\_p-j} p^{M\_p-j-1} \equiv 2(mod \mathcal{M}) \tag{29}$$

**LEMMA 2.** If *Fn* fermat number and *<sup>p</sup>* <sup>¼</sup> <sup>2</sup>*nFn* <sup>þ</sup> 1 where *<sup>p</sup>* is a prime then

$$\sum\_{j=1}^{F\_n-2} \pi\_j (\mathfrak{2}^n)^{F\_n-j-1} \equiv \mathfrak{2} (\mathfrak{2}^n)^{F\_n} (mod \, p) \tag{30}$$

**Proof.** Let be in theorem 1 *<sup>b</sup>* <sup>¼</sup> <sup>1</sup> *and a* <sup>¼</sup> <sup>2</sup> *and m* <sup>¼</sup> <sup>2</sup>*<sup>n</sup>* then we get *<sup>λ</sup>* <sup>¼</sup> *am* <sup>þ</sup> *<sup>b</sup>* <sup>¼</sup> 22*n* <sup>þ</sup> 1 and <sup>M</sup> <sup>¼</sup> *ma<sup>m</sup>* <sup>þ</sup> *bm* <sup>þ</sup> <sup>1</sup> <sup>¼</sup> <sup>2</sup><sup>2</sup>*n*þ*<sup>n</sup>* <sup>þ</sup> <sup>2</sup>*<sup>n</sup>* <sup>þ</sup> <sup>1</sup> <sup>¼</sup> *<sup>p</sup>* where

$$\sum\_{j=1}^{F\_n-2} \pi\_j (\mathfrak{L}^n)^{F\_n-j-1} \equiv \mathfrak{L} (\mathfrak{L}^n)^{F\_n} (mod \ p) \tag{31}$$

**REMARK 2**: Fermat,s numbers *Fn* <sup>¼</sup> 22*<sup>n</sup>* þ 1 are named after pierre de farmat because he was the first to stud these numbers guess that all fermat numbers are prime

$$\{3, 5, 17, 57, 6557, \dots\} \tag{32}$$

But this conjecture was denied by Euler's proved the Fermat number *F*<sup>5</sup> is not prime

$$F\_5 = 2^{2^5} + 1 = 4294967297 = 641 \times 6700417 \tag{33}$$

These numbers was named 2*<sup>P</sup>* � 1 Mersnne numbers, so in Ref. to Marin Meresenne, who began studying them by 2020 he discovered fifty –one prime numbers. There is a program called (the big search for Mersenne prime on the internet). Many prime numbers of Meresnne numbers have been discovered, we know about M2, M3, M5, M7, M17, M19, M31, M521 … , M1279, M110305, M132049, M25964951 all prime numbers *M*<sup>11</sup> is not prime number and they give good results from fermat numbers that only four digits of it have been discovered so far. We know about Fermat numbers, if *Fn* is not prime, then there is *<sup>b</sup>* <sup>¼</sup> *<sup>k</sup>*2*<sup>n</sup>*þ<sup>2</sup> <sup>þ</sup> 1 where *Fn*∕*b*, and likewise abuot Mersenne numbers, if *Mp* is not prime, there is *q* ¼ 2*pr* þ 1 where *Mp*∕*q* and q is prime. From a computational point of view, we find that the results that we have reached are more robust and generalizable those of the results mentioned. Firstly, this is due to the existing ideas and properties is those results. This is represented in highlighting an integral relationship between two prime numbers and more prime numbers. We notice in the LEMMA 1 that the prime numbers and the *<sup>p</sup>* <sup>¼</sup> <sup>3</sup>*<sup>m</sup>* � 2 and the number in form <sup>M</sup> <sup>¼</sup> *<sup>m</sup>*3*<sup>m</sup>* � <sup>2</sup>*<sup>m</sup>* <sup>þ</sup> 1 numbers in form combine in one result, and also the properties of the Mersnne numbers. Such a correlation does not exist [5, 6] as well as with the ratio of the Fermat numbers also meet with the numbers in LEMMA 2 and this shows relationship between the Fermat numbers and those numbers. In addition to that, the result are expressed a polynomial that highlights the properties of those numbers, and it can also be used as a primitive test to discover those numbers . For a discussion of such issues see [3, 8, 9, 11–14] on there are several numbers studied *<sup>A</sup>*3*<sup>n</sup>* � 1, *<sup>k</sup>*2*<sup>n</sup>* <sup>þ</sup> 1, 2*<sup>n</sup>* � 1 and close to these formulas.
