Some Extensions on Numbers

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1. 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 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” [1], we know prime numbers and primality test. From this paper “Fermat’s little theorem” [2], we know various types of explanations about Fermat’s little theorem.

Prepositions 2 to 6 are worked by me. They are noted as PRB which means Prema. R. Balasubramani [3]. They are published in Fermat’s theorem one extension: Mathematical Sciences International Journal ISSN 2278-8697 VOLUME 8 ISSUE 1 (JUNE 2019), P. 6-10.

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.

2. Discussion and Results

Hint 1: Define. Then .

Hint 2: Define. Then .

2.1. Let’s Now Examine φ(pn) When p Is a Factor of n

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 and p divides n the following result follows: if and only if for any natural number r.

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

For our example we choose 20, so let’s consider 2 × 20; . Putting together the two sets mentioned in our previous examples we have {1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39}, exactly all 16 numbers are relatively prime to 40.

2.2. 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. Notice the conditions imply iff and.

Suppose the list of multiples is, 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.□

For our examples we choose 20, so let’s consider 3 × 20;. Putting together the two sets mentioned in our previous examples we have {1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59} exactly all 16 numbers relatively prime to 60.

Preposition 1: Let n be a positive integer. Then when n is composite number and when n is prime number.

Proof: Let n be a positive integer.

When n be a composite number and n divides, then .

Notice that all the numbers that are relatively prime to are also relatively prime to. Since And n divides The following result follows: if and only if for any natural number r.

There are n intervals, each with numbers relatively prime to, hence by the hint 1: the set has elements.

When n be a prime and n does not divide, then .

We know that is the number of numbers relatively prime to and less than. Notice that all the multiples of n whose factors are relatively prime to are counted, since. Notice the conditions imply if and only if and.

Suppose the list of multiples is, where all the r’s are relatively prime to. The set has numbers relatively prime to and 0 relatively prime to n, because they are all multiples of n. by this way we get.

Preposition 2 (PRB): Let n be a positive integer. Then where n_{i}’s are composite numbers and’s are prime numbers not exceeding n.

Proof:

Using preposition 1, we obtained when n is composite number and

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

(1)

Example 1: Find the value of

Solution:

Let then we can write. So

.

Example 2: Find the value of

Solution:

Let then. So

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 integers less than n^{a} that are relatively prime to n^{a}.

Hence,

(2)

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.

2.3. 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”, and n be an exponent to “a” then. 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 is any exponent.

Proof: Let p be a prime, and a is a positive integer with p does not divides a, and n be an exponent to a then

;

if then

if then

Do this again and again until we get

if.

Hence we get,

(3)

2.4. 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 then.

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

(4)

Proof:

Let. So we can write for some integer m. now we can write. Here, since k value has φ(m) as a one factor and n is a positive integer.

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

(5)

Proof: Let then. So we can write for some integer m. Now. Here. Since k has φ(m) as a one factor and h is a positive integer. It gives. □

2.5. Exponent Parallelogram

Definition 1: Let and be the exponent to m then do 1^{st} operation is subtracting each element with its successive element of 1^{st} line elements. Result will be, we shall name as “a”. 2^{nd} operation is subtracting each element with its successive element of 1^{st} operation, result will be then we shall name as “a^{2}”. 3^{rd} operation is subtracting each element with its successive element of 2^{nd} operation, result will be then we shall name as “a^{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: should be placed between and in kth operation. Because.

Let we construct exponent plane for 5: for and

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, where “k” is any positive integer of 1^{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,) and with (4, 7) = 1, then.

2) Let 7 is a first line prime exponents, (6, 42, 294, 2058,) are 1^{st} operation to kth operation and with (4, 7) = 1, then. Where 6, 42, 294, 2058, are coherence numbers of φ(7).

3) Let 5 is a first line prime exponents, (4, 5 × 4, 25 × 4, 125 × 4, , 16, 5 × 16, 25 × 16, , 64, 5 × 64, 25 × 64,) are 1^{st} operation to kth operation and with (4, 5) = 1, then, , , ,. Where (4, 5 × 4, 25 × 4, 125 × 4, , 16, 5 × 16, 25 × 16, , 64, 5 × 64, 25 × 64,) are coherence numbers of φ(5) and.

2.6. Prime Bases on Fermat’s Last Theorem

Let we see following summations.

Let are prime numbers then

;

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

(6).

Theorem 1: Let are prime numbers then. Where q is any prime.

Proof:

Let then

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

Case 2: If P is composite, we can write. if then result is obvious.

Case 3: If P is composite and, then we can write. If are distinct primes then result is obvious. But if we get. This result contradict with (1). So. Where q is any prime.

2.7. Primality of Pascal’s Triangle

Definition 2: For all and for all reals we have the formula. For every we have then p divides and every; where is called Primality of binomial expansion.

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 where

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.

2.8. Constructing Pascal’s Triangle by Arithmetic Triangles

Addition triangle

Definition 3: Let 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:

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

- - - - - - - -

- - - .

2.9. Backward Difference Triangle

Definition 4: Let then do 1^{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

- - - - - - - -

- - - , where sign depends upon whether n is odd or even. If n is odd we get, else we get.

2.10. Forward Difference Triangle

Definition 5: Let 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:

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

- - - - - - - -

- - - where sign depends upon whether n is odd or even. If n is odd we get, else we get.

2.11. Multiplication Triangle

Definition 6: Let then do 1^{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

- - - - - - - -

- - -

2.12. Forward Division Triangle

Definition 7: Let 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 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

- - - - - - - -

- - - , where sign depends upon whether n is odd or even. If n is odd we get, else we get.

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

2.13. Backward Division Triangle

Definition 8: Let 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:

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

- - - - - - - -

- - - , where sign depends

2.14. Backward Exponent Difference Triangle

Definition 9: Let then do 1^{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 be an exponent of any then “n”th difference of would be n!.

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:. □

2.15. Forward Exponent Difference Triangle

Definition 10: Let 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 degrees construct Pascal’s triangle.

Theorem 3: Let be an exponent of any then “n”th difference of would be.

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

References

[1] Pomerance, C. and Crandall, R. (2005) Prime Numbers: A Computational Perspective. 2nd Edition, Springer, New York.

[2] Chaudhuri, S. (2014) Fermat’s Little Theorem-CS 2800: Discrete Structures.

[3] Balasubramani, P.R. (2019) Fermat’s Theorem One Extension. Mathematical Sciences International Journal, 8, 6-10.