**3. Prime numbers In form** *p* ¼ *am* þ **1**

In this section we study the prime numbers in form *p* ¼ *am* þ 1 *where a*, *m* >1∈ which is a special case of the prime numbers form *<sup>a</sup><sup>m</sup>* <sup>þ</sup> *bm* <sup>þ</sup> 1, when substituting *<sup>b</sup>* <sup>a</sup> certain value, we study the properties of numbers *p* ¼ *am* þ 1, *a*, *m* ∈ and *q* ¼ *bp* þ 1 *where b*, *p* >1∈ , as well as relationship between polynomial. These polynomial numbers show us special properties of this numbers *am* þ 1 as well as the numbers *p* ¼ *qa* þ 1 *where q is a prime number* of the properties, polynomial can be used as a primitive testing algorithm for those numbers. The proof depend mainly on THEOREM 1. In this section, we explain and realize that there is more than one variable in the theorem, for example, *ma<sup>m</sup>* <sup>þ</sup> *bm* <sup>þ</sup> 1 variable *m*, *a*> 1∈ *and b*∈, *b* 6¼ 0, because of this, have many distinctive properties. We prove THEOREM 3 is this section and THEOREM 4 by directly changing the value of that variable without any the complexity mentions in particular the basic operations and the binomial theorem are the other extreme in the proofs.

**THEOREM 3.** *if p* ¼ *qm* þ 1 where p, q *is a prime odd* and *a*, *m* > 1∈ where *Cm*ð Þ¼ *<sup>a</sup> ma<sup>m</sup>* <sup>þ</sup> 1 and *<sup>π</sup><sup>j</sup>* <sup>¼</sup> <sup>1</sup> *q q j* � � then X *q*�2 *j*¼1 *<sup>π</sup>j*ð Þ �*Cm*ð Þ *<sup>a</sup> <sup>q</sup>*�*j*�<sup>1</sup> *<sup>q</sup>* � *<sup>a</sup><sup>m</sup>* ð Þ� *<sup>χ</sup>*ð Þ *<sup>m</sup>*,*q*�*am* ð Þ *mod p* (34)

**Proof.** let be in theorem *<sup>b</sup>* <sup>¼</sup> *<sup>q</sup>* � *<sup>a</sup><sup>m</sup> where q* <sup>&</sup>gt;<sup>2</sup> *is prime odd* then <sup>M</sup> <sup>¼</sup> *ma<sup>m</sup>* <sup>þ</sup> *<sup>q</sup>* � *<sup>a</sup><sup>m</sup>* ð Þ*<sup>m</sup>* <sup>þ</sup> <sup>1</sup> <sup>¼</sup> *qm* <sup>þ</sup> <sup>1</sup> <sup>¼</sup> *<sup>p</sup>* therefore also *<sup>λ</sup>* <sup>¼</sup> *<sup>a</sup><sup>m</sup>* <sup>þ</sup> *<sup>b</sup>* <sup>¼</sup> *<sup>a</sup><sup>m</sup>* <sup>þ</sup> *<sup>q</sup>* � *<sup>a</sup><sup>m</sup>* <sup>¼</sup> *<sup>q</sup>* so we have that

$$\eta\_{(\lambda)}(m,\mathbf{q}-a^m) = \sum\_{j=1}^{q-2} \pi\_j (q - a^m)^{q-j} m^{q-j-1} \tag{35}$$

And

$$\chi\_{(m,q^{-a\mathfrak{m}})} = (mq - ma^m)^\dot{\lambda} - (-m)^\dot{\lambda} - q + a^m + \mathbf{1} \tag{36}$$

Note that *m q* � *<sup>a</sup><sup>m</sup>* ð Þ¼ *mq* � *ma<sup>m</sup>* <sup>¼</sup> *<sup>p</sup>* � *ma<sup>m</sup>* � <sup>1</sup> *le be Cm*ð Þ¼ *<sup>a</sup> mam* <sup>þ</sup> 1 then we have

$$\eta\_{(q)}(m, q - a^m) = \sum\_{j=1}^{q-2} \pi\_j (p + (-\mathbb{C}\_m(a)))^{q-j-1} (q - a^m) \tag{37}$$

Note that from binomial theorem

$$(p + (-\mathcal{C}\_m(a)))^{q^\*-j-1}$$

$$=\sum\_{k=0}^{q-j-2} \binom{q-j-1}{k} p^{q-j-1-k} (-\mathcal{C}\_m(a))^k (q - a^m)$$

$$+(-\mathcal{C}\_m(a))^{q^\*-j-1} (q - a^m) \text{ for all } 1 \le j \le q - 1 \tag{38}$$

$$y\_m = \sum\_{k=1}^{q-j-2} \binom{q-j-1}{k} p^{q-j-1-k} (-\mathcal{C}\_m(a))^k \tag{39}$$

$$\eta\_{(q)}(m, q - a^m) = \sum\_{j=1}^{q-2} \pi\_j \wp\_m + \sum\_{j=1}^{q-2} \pi\_j (-\mathcal{C}\_m(a))^{q-j-1} (q - a^m) \tag{40}$$

$$\sum\_{j=1}^{q-2} \pi\_j \wp\_m \equiv 0 (mod \ p) \text{ all } j = 1, 2, 3 \dots q - 1 \tag{41}$$

$$
\eta\_{(q)}(m, q - a^m) \equiv \chi\_{(m, q - a^m)}(mod \ p) \tag{42}
$$

$$\sum\_{j=1}^{q-2} \pi\_j (-\mathcal{C}\_m(a))^{q-j-1} (q - a^m) \equiv \chi\_{(m, q-a^m)}(mod \ p) \tag{43}$$

$$\begin{aligned} \chi\_{(mq - a^m)} &= (mq - ma^m)^q - (-m)^q + \mathbf{1} - q + a^m \\ &= (p + (-ma^m - \mathbf{1}))^q - (-m)^q - q + a^m + \mathbf{1} \\ &= \sum\_{j=0}^{q-1} \binom{q}{j} p^{q-j} (-ma^m - \mathbf{1})^j + (-ma^m - \mathbf{1})^q - (-m)^q - q + a^m + \mathbf{1} \end{aligned} \tag{44}$$

$$\sum\_{j=0}^{q-1} \binom{q}{j} p^{q-j} (-ma^m - 1)^j \equiv 0 (mod \ p) \tag{45}$$

$$\chi\_{(m,q-a^m)} = (-ma^m - 1)^q - (-m)^q - q + a^m + 1 \tag{46}$$

Therefore

$$\sum\_{j=1}^{q-2} \pi\_j (-\mathcal{C}\_m(a))^{q-j-1} (q - a^m) \equiv \chi\_{(m, q-a^m)}(mod \ p) \tag{47}$$

**LEMMA 3**. *if p* ¼ 2*q* þ 1 and *q is a prime odd* then

$$\sum\_{j=1}^{q-2} \pi\_j \left(-2a^2 - 1\right)^{q-j-1} (q - a^m) \equiv \chi\_{(2, q-a^2)}(mod \, p) \tag{48}$$

**Proof.** Let be in theorem 3 *m* ¼ 2

**LEMMA 4**. *if p* ¼ *qm* þ 1 *and q is prime odd and m* >1∈ then

$$\sum\_{j=1}^{q-2} \pi\_j (-m2^m - 1)^{q-j-1} (q - 2^m) \equiv \chi\_{(m, q-2^m)}(mod \ p) \tag{49}$$

**Proof.** Le be in theorem 3 *a* ¼ 2

**REMARK 3:** if we make a comparison between the results found in [2, 3] about the generalized Cullen numbers and the results that we reached here, in fact, we find that these results are more generalized than those, and also rich in properties than those. The first general and the Cullen numbers in particular, and this is considered one of the properties of the prime numbers of the number in an adjective, as well as this relationship in the form a polynomial that combines the Cullen numbers and the prime number in general *P* ¼ *qa* þ 1 *where q is prime odd note P*∈ � f g 2, 3, 5 . Such ideas do not exist in 5, 6, 7, ½ � … *:*12 . we also note that polynomials can be used as a primitive test to discover prime numbers.

**THEOREM 4.** *if q* >2, *m* >1∈ *and p* ¼ *qm* þ 1 and p is prime then

$$\sum\_{j=1}^{q-2} \binom{q}{j} (-\mathcal{C}\_m(a))^{q-j-1} (q - a^m) \equiv \chi\_{(q, m, q-a^m)}(mod \ p) \tag{50}$$

**Proof.** we will prove this theorem with the same ideas as in the proof THEOREM 3 now according THEOREM 1 we have that

$$\eta\_{(\lambda)}(b,m) = \sum\_{j=1}^{\lambda-2} \binom{\lambda}{j} b^{\lambda-j} m^{\lambda-j-1} \tag{51}$$

And

$$
\chi\_{(\lambda, b, m)} = \lambda (bm)^{\dot{\lambda}} - \lambda (-m)^{\dot{\lambda}} + \dot{\lambda} - b\dot{\lambda} \tag{52}
$$

Then let be *<sup>b</sup>* <sup>¼</sup> *<sup>q</sup>* � *<sup>a</sup><sup>m</sup> and m*, *<sup>a</sup>*, *<sup>q</sup>*>1<sup>∈</sup> where *<sup>λ</sup>* <sup>¼</sup> *<sup>a</sup><sup>m</sup>* <sup>þ</sup> *<sup>b</sup>* <sup>¼</sup> *<sup>a</sup><sup>m</sup>* <sup>þ</sup> *<sup>q</sup>* � *<sup>a</sup><sup>m</sup>* <sup>¼</sup> *<sup>q</sup>* then we have that

$$\eta\_{(q)}(q - a^m, m) = \sum\_{j=1}^{q-2} \binom{q}{j} (q - a^m)^{q-j} m^{q-j-1}$$

$$= \sum\_{j=1}^{q-2} \binom{q}{j} (p + (-\mathcal{C}\_m(a)))^{q-j-1} (q - a^m) \tag{53}$$

$$\eta\_{(q)}(q - a^m, m) = \sum\_{j=1}^{q-2} \binom{q}{j} \sum\_{k=0}^{q-j-2} \binom{q-j-1}{j} p^{q-j-1-k} (-\mathcal{C}\_m(a))^k (q - a^m)$$

$$+ \sum\_{j=1}^{q-2} \binom{q}{j} (-\mathcal{C}\_m(a))^{q-j-1} (q - a^m) \tag{54}$$

$$\sum\_{j=1}^{q-2} \binom{q}{j} \sum\_{k=0}^{q-j-2} \binom{q-j-1}{j} p^{q-j-1-k} (-\mathcal{C}\_m(a))^k (q-a^m) \equiv \mathbf{0}(mod \ p) \tag{55}$$

$$
\eta\_{(q)}(q - a^m, m) \equiv \chi\_{(\, \, q, q - a^m, m)}(mod \, p) \tag{56}
$$

$$\sum\_{j=1}^{q-2} \binom{q}{j} (-\mathcal{C}\_m(a))^{q-j-1} (q - a^m) \equiv \chi\_{(\;q, q-a^m, m)}(mod \; p) \tag{57}$$

$$\chi\_{(q,q-a^m,m)} = q(qm - ma^m)^q - q(-m)^q + q - q^2 + qa^m \tag{58}$$

$$q(qm - ma^m)^q = q \sum\_{j=0}^{q-1} \binom{q}{j} p^{q-j} (-ma^m - 1)^j + q(-ma^m - 1)^q \tag{59}$$

$$\sum\_{j=0}^{q-1} \binom{q}{j} p^{q-j} (-ma^m - 1)^j \equiv \mathbf{0} (mod \ p) \tag{60}$$

$$\chi\_{(q,q-a^m,m)} = q(-ma^m-1)^q - q(-m)^q + q - q^2 + qa^m \tag{61}$$

Then

$$\sum\_{j=1}^{q-2} \binom{q}{j} (-\mathcal{C}\_m(a))^{q-j-1} (q-a^m) \equiv \chi\_{(\;q,q-a^m,m)}(mod\;p) \tag{62}$$

**LEMMA 5**. *if p* ¼ 6*m* þ 1 and *a*>1∈ then

$$\sum\_{j=1}^{4} \binom{6}{j} (-\mathcal{C}\_m(a))^{5-j} (6-a^m) \equiv \chi\_{(6,m,6-a^m)}(mod\ p) \tag{63}$$

**Proof.** Let be in theorem 4 *q* ¼ 6
