Some Extensions on Numbers

My previous work dealt finding numbers which relatively prime to factorial value of certain number, high exponents and also find the way for finding mod values on certain number’s exponents. Firstly, I retreat my previous works about Euler’s phi function and some works on Fermat’s little theorem. Next, I construct exponent parallelogram to find coherence numbers of Euler’s phi functioned numbers and apply to Fermat’s little theorem. Then, I test the primality of prime numbers on Pascal’s triangle and explore new ways to construct Pascal’s triangle. Finally, I find the factorial value for certain number by using exponent triangle.


Introduction
We know Fermat's little theorem and Euler's φ (phi) function. Such are well defined operations on number theory and algebra. Euler's φ (phi) function is considered as general proof of Fermat's little theorem. We seek other ways to find mod values on Fermat's little theorem, and generalize φ (phi) function for a certain integer's exponentiation and factorial value. We construct the exponent parallelogram to find the coherence values of Euler's φ (phi) function. We find higher valued exponents on Fermat's little theorem according to this. We also specify Fermat's last theorem by using prime numbers. Also we know binomial coefficients are constructing Pascal's triangle, in which we see the divisibility of prime numbers (primality test) in prime number exponentiation on Pascal's triangle. In addition, we construct Pascal's triangle and seek other ways except for binomial coefficients, i.e. and construct Pascal's triangle by arithmetic operations triangle. Finally instead of binomial coefficients in Pascal' triangle, we use exponents value of certain integer to construct Pascal's triangle, and then use "n"th expansion to find factorial of such certain number.
First Blaise Pascal (1623-1662) introduced Pascal's triangle, after that, Isaac Newton (1643-1727) used the facts of Pascal's triangle he developed binomial expansion. He and his followers used binomial theorem for Probability and Statistical problems. Factorial were used to count permutations at as early as the 12 th century, by Indian scholars. In 1677, Fabian Stedman described factorial as applied to change ringing, a musical art involving the ringing of many tuned bells. In his words "Now the nature of these methods is such that the change of one number comprehends (includes) changes on lesser numbers". In that mean period, James Stirling (1692-1770) first introduced one approximation for finding nth factorial of a certain number. Then Adrien-Marrie Legendre used Leonhard Euler's (1707-1783) second integral formula and notated a symbol for it and then named it as Gamma function. It was a good approximation finding factorial of Real numbers. Jacques Philippe Marie Binet (1786-1856), modified James Stirling's approximation. Finally, the notation n! was introduced by the French mathematician Christian Kramp in 1808. Pierre

Let's Now Examine φ(pn) When p Is Not a Factor of n
Lemma 2: Let p be a prime and p does not divide n, then Proof: We know that pΦ(n) is the number of numbers relatively prime to n and less than pn. Notice that all the multiples of p whose factors are relatively prime to n are counted, since , where all the r's are relatively prime to n. the set has Φ(n) numbers relatively prime to n and 0 relatively prime to p, because they are all multiples of p. we subtract this many from our original count and we have  When n be a composite number and n divides ( ) Notice that all the numbers that are relatively prime to ( ) When n be a prime and n does not divide ( ) is the number of numbers relatively prime to ( ) 1 ! n − and less than ( ) 1 ! n n − . Notice that all the multiples of n whose factors are relatively prime to ( ) Suppose the list of multiples is  when n is prime number. Since all even numbers are composites except 2 because 2 is prime. So we cannot find an even composite number less than four. And two is the only prime number less than three. Also 1 is the only number relatively prime to two and below it. So we obtained from these two equations we get Example 1: Find the value of ( ) Example 2: Find the value of ( ) Proof: The positive integers less than n a that are not relatively prime to n are those integers not exceeding n a that are divisible by n. There are exactly n a-1 such integers, so there are ( ) 1 a n n ϕ − integers less than n a that are relatively prime to n a .

Exponent Division on Fermat's Little Theorem
Preposition 4 (PRB): If p is prime and "a" is a positive integer with p does not divides "a", . r is a congruent of "a" for mod p, where "s" is a quotient and "t" is a residue when "n" divided by p and n ∈ N is any exponent.
Proof: Let p be a prime, and a is a positive integer with p does not divides a,

Proving Fermat's Little Theorem, Using Preposition 4
If p is prime and a is a positive integer with p does not divides "a" and    By the above results we define, 1) If "E" is a 1 st line prime exponent and "a" is an integer with (a, E) = 1, then 2) If "E" is a prime exponent and "a" is an integer with (a, E) = 1, then , where "k" is any positive integer of 1 st operation to k-th operation coherence numbers of φ(E).

Prime Bases on Fermat's Last Theorem
Let we see following summations.
Let i p are prime numbers then From the above recursion, we formulate the result then we get, ( )

Constructing Pascal's Triangle by Arithmetic Triangles
Addition triangle Definition 3: Let , , , A B C Z ∈  then do 1 st operation is adding each element with its successive element of 1 st line elements, 2 nd operation is adding each element with its successive element of 1 st operation, and 3rd operation is adding each element with its successive element of 2nd operation. By this way we do the same up to nth operation. These 1 st line to nth operation diagonal elements coefficients construct Pascal's triangle. Now we construct addition triangle:

Backward Difference Triangle
where 0 n       sign depends upon whether n is odd or even. If n is odd we get

Forward Difference Triangle
Definition 5: Let , , , A B C Z ∈  then do 1 st operation is subtracting each element with its successive element of 1 st line elements, 2 nd operation is subtracting each element with its successive element of 1 st operation, and 3rd operation is subtracting each element with its successive element of 2nd operation. By this way we do the same up to nth operation. These 1 st line to nth operation diagonal elements coefficients construct Pascal's triangle with negative coefficients.
Now we construct forward difference triangle: From the above, using the colored diagonal we can construct a negative Pascal's triangle: where n k       sign depends upon whether n is odd or even. If n is odd we get n k   −     , else we get n k       .

Forward Division Triangle
where n k       sign depends upon whether n is odd or even. If n is odd we get n k   −     , else we get n k Upon whether n is odd or even. If n is odd we get n k       , else we get n k   −     .

Backward Division Triangle
Definition 8: Let , , , A B C Z ∈  then do 1 st operation is dividing each element with its successive element of 1 st line elements, 2 nd operation is dividing each element with its successive element of 1 st operation, and 3rd operation is dividing each element with its successive element of 2nd operation. By this way we do the same up to nth operation. These 1 st line to nth operation diagonal elements degrees construct Pascal's triangle. Now we construct backward division triangle:

Conflicts of Interest
The author declares no conflicts of interest regarding the publication of this paper.