_{1}

^{*}

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.

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 de Fermat (1601-1665) stated Little theorem, for any two relatively prime numbers, in which exponent should be prime number; after that Leonhard Euler (1707-1783) found Totient function and then generalized Fermat’s little theorem for any two relatively prime numbers.

From this book “Prime numbers a computational Perspective” [

Prepositions 2 to 6 are worked by me. They are noted as PRB which means Prema. R. Balasubramani [

In this paper,

1) I retreat my previous work Fermat’s theorem one extension. Here I extend my works to finding the coherence numbers (constructing exponent parallelogram) for Euler’s phi function and then generalize it for Fermat’s little theorem.

2) I test the primality of prime numbers on Pascal’s triangle.

3) I specify Fermat’s last theorem by prime numbers.

4) I generate Pascal’s triangle by arithmetic operations.

5) I find factorial value for certain number by using exponent triangle.

Hint 1: Define

Hint 2: Define

Lemma 1: Let p be a prime and p divides n, then

Proof: Notice that all the numbers that are relatively prime to pn are also relatively prime to n. since

There are p intervals, each with Φ(n) numbers relatively prime to pn, hence by the hint 1: the set

For our example we choose 20, so let’s consider 2 × 20;

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

Suppose the list of multiples is

For our examples we choose 20, so let’s consider 3 × 20;

Preposition 1: Let n be a positive integer. Then

Proof: Let n be a positive integer.

When n be a composite number and n divides

Notice that all the numbers that are relatively prime to

There are n intervals, each with

When n be a prime and n does not divide

We know that

Suppose the list of multiples is

Preposition 2 (PRB): Let n be a positive integer. Then _{i}’s are composite numbers and

Proof:

Using preposition 1, we obtained

Example 1: Find the value of

Solution:

Let

Example 2: Find the value of

Solution:

Let

Preposition 3 (PRB): Let n be a positive integer and “a” be an exponent to n. Then

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 ^{a} that are relatively prime to n^{a}.

Hence,

Example 4: Find the value of φ(10^{4}).

Solution:

Let φ(10^{4}) then

Since 10 is a composite, 10^{4} = 10,000 so φ(10,000) = 4000.

Example 5: Find the value of φ(331^{5}).

Solution:

Let φ(331^{5}) then

Since 331 is a prime, 331^{5} = 3,973,195,810,651 so φ(3,973,195,810,651) = 3,961,192,197,930.

Preposition 4 (PRB): If p is prime and “a” is a positive integer with p does not divides “a”,

Proof: Let p be a prime, and a is a positive integer with p does not divides a,

Do this again and again until we get

Hence we get,

If p is prime and a is a positive integer with p does not divides “a” and

ð

Example 6: Find the value of 3^{1900} mod 13.

Solution: We can write

1900 = 146.13 + 2

≡146 + 2 = 148 here 148 ≥ 13 so,

148 = 11.13 + 5

≡16 here 16 ≥ 13 so,

19 = 1.13 + 3

≡4 here 4 < 13 so,

Apply this algorithm, then we get

Preposition 5 (PRB): If m is a positive integer and a is an integer with (a, m) = 1,

Then

where

Proof:

Let

It gives

Example 8: Find the value of

Solution:

Preposition 6 (PRB): If m is a positive integer and a is an integer with (a, m) = 1,

Then

where

Proof: Let

Definition 1: Let ^{st} operation is subtracting each element with its successive element of 1^{st} line elements. Result will be^{nd} operation is subtracting each element with its successive element of 1^{st} operation, result will be ^{2}”. 3^{rd} operation is subtracting each element with its successive element of 2^{nd} operation, result will be ^{3}”. By this way we do the same up to nth operation. These 1^{st} line to nth operation diagonal elements coefficients construct exponent parallelogram.

Now we construct exponent parallelogram:

Note:

Let we construct exponent plane for 5: for

Now we get,

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^{st} operation to k-th operation coherence numbers of φ(E).

Examples:

1) Let 7 is a first line prime exponents i.e. (1, 7, 49, 343,

2) Let 7 is a first line prime exponents, (6, 42, 294, 2058,^{st} operation to kth operation and

3) Let 5 is a first line prime exponents, (4, 5 × 4, 25 × 4, 125 × 4, ^{st} operation to kth operation and

Let we see following summations.

Let

For squared primes:

For cubed primes:

For fourth exponent primes:

By this way we concluded,

where

From the above recursion, we formulate the result then we get,

where

Theorem 1: Let

Proof:

Let

Case 1: If P is prime, result is obvious.

Case 2: If P is composite, we can write

Case 3: If P is composite and

Definition 2: For all

Prime number Pascal’s triangle coefficients

0 1

1 1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

5 1 5 10 10 5 1

6 1 6 15 20 15 6 1

7 1 7 21 35 35 21 7 1

…

11 1 11 55 165 330 462 462 330 … 1

…

p

Examples:

1) 7 divides 7 + 21 + 35 + 35 + 21 + 7 i.e. 126/7 = 18

2) 11 divides 2(11 + 55 + 165 + 330 + 462) i.e. 2046/11 = 186.

Addition triangle

Definition 3: Let ^{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:

1^{st} line:

1^{st} operation:

2^{nd} operation:

3^{rd} operation:

4^{th} operation:

5^{th} operation:

From the above, using the colored diagonal we can construct a Pascal’s triangle:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

- - - - - - - -

Definition 4: Let ^{st} operation is subtracting each element with its predecessor element of 1^{st} line elements, 2^{nd} operation is subtracting each element with its predecessor element of 1^{st} operation, and 3rd operation is subtracting each element with its predecessor 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 backward difference triangle:

1^{st} line:

1^{st} operation:

2^{nd} operation:

3^{rd} operation:

4^{th} operation:

5^{th} operation:

From the above, using the colored diagonal we can construct a negative Pascal’s triangle:

1

−1 1

1 −2 1

−1 3 −3 1

1 −4 6 −4 1

−1 5 −10 10 −5 1

1 −6 15 −20 15 −6 1

- - - - - - - -

Definition 5: Let ^{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:

1

1 −1

1 −2 1

1 −3 3 −1

1 −4 6 −4 1

1 −5 10 −10 5 −1

1 −6 15 −20 15 −6 1

- - - - - - - -

Definition 6: Let ^{st} operation is multiplying each element with its successive element of 1^{st} line elements, 2^{nd} operation is multiplying each element with its successive element of 1^{st} operation, and 3rd operation is multiplying 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 multiplication triangle:

From the above, using the colored diagonal exponents, we can construct a Pascal’s triangle:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

- - - - - - - -

Definition 7: Let ^{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 forward division triangle:

From the above, using the colored diagonal exponents, we can construct a Pascal’s triangle:

1

1 −1

1 −2 1

1 −3 3 −1

1 −4 6 −4 1

1 −5 10 −10 5 −1

1 −6 15 −20 15 −6 1

- - - - - - - -

Upon whether n is odd or even. If n is odd we get

Definition 8: Let ^{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:

From the above, using the colored diagonal exponents, we can construct a Pascal’s triangle:

1

−1 1

−1 2 −1

−1 3 −3 1

−1 4 −6 4 −1

−1 5 −10 10 −5 1

−1 6 −15 20 −15 6 −1

- - - - - - - -

Definition 9: Let ^{st} operation is multiplying each element with its successive element of 1^{st} line elements, 2^{nd} operation is multiplying each element with its successive element of 1^{st} operation, and 3rd operation is multiplying 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.

Theorem 2: Let

Let we construct backward difference triangle, in which first line numbers are “n”th exponent of whole numbers. For any

1^{st} line:

1^{st} operation:

2^{nd} operation:

3^{rd} operation:

…

nth operation:

Definition 10: Let ^{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 degrees construct Pascal’s triangle.

Theorem 3: Let

Proof:

Let we construct forward difference triangle, in which first line numbers are “n”th exponent of whole numbers. For any

1^{st} line:

1^{st} operation:

2^{nd} operation:

3^{rd} operation:

…

nth operation:

Examples for backward exponent difference method:

1) Let m = 0 and n = 5 then

2) Let m = −1 and n = 5 then

3) Let m = 1 and n = 5 then

Examples for backward exponent difference method:

1) Let m = 0 and n = 4 then

2) Let m = −1 and n = 4 then

3) Let m = 1 and n = 5 then

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

Rangasamy, B.P. (2019) Some Extensions on Numbers. Advances in Pure Mathematics, 9, 944-958. https://doi.org/10.4236/apm.2019.911047