CHAPTER 21

† The newest of methods of solving††† and calculating

 

Appendixes (I)

 

 

 

 

 

 

Contents

 

Ÿ Solving congruence and Diophantine equations by "H.M" table (New Method):

†††††††††††††††††††††††††††††††††††††††††††

Ÿ Solving Diophantine equation of order in by "H.M" table:

†††††††††††††††††††††††††††††††††††††

Ÿ A new and fast method for calculating "" determinant (H.M method)

Ÿ Definition of regular and irregular prime numbers by "H.M" determinant

Ÿ New method of calculation of sum of "k-th" power of the first "n" natural numbers by "H.M" determinant (expressing "" via a determinant)

Ÿ Determining the number of roots of perfect cubic degree of equation directly by "H.M" method

Ÿ Proof of a new and applied "H.M" theorem (concerning the factorization of composite numbers)


21.1. Solving congruence and Diophantine equations
by "H.M" table (new method)†

†††††††† ††††(1)

We call number "n" as module, "x" as unknown, "a" as coefficient of unknown and "c" as constant number of congruence.

We know that if "" and according to multiplicative inverse of "a" namely †(multiplicative inverse of "a" in congruence):

Equation (1) can be written in below form:

Therefore, for solving equation (1), it is enough to calculate inverse of "a". We know that inverse of "a", (multiplicative inverse of "a") is calculated from below congruence equation:

††††††††††† (2)

In equation (2) if †and †is dividing residue of , we can write:

† †††††††(3)

Congruence equation (3) can be written in below Diophantine equation form:

†††††††††† (4)

According to "" and if †is dividing residue of , we can write:

††††††††††† (5)

As previous stage, congruence equation (5) can be written in below form:

†††††††††††††† (6)

As the same method, If we continue this action, coefficient of unknown will be "1" and with return to previous stages, we can calculate "x" with a reversible method.

21.1.1. Attention.

We calculate all of above calculations with "4- row" table. This table has "4" rows that is completed after "4" stages.

For example we solve below congruence equation:

By using "H.M" table:

 

1378

621

136

77

59

18

5

3

2

1

1

-1

1

-1

1

-1

1

-1

1

 

621

136

77

59

18

5

3

2

1

 

537

242

53

30

23

7

2

1

1

 

21.1.2. Attention

Every table column, determines a congruence equation.

 

In this example, since, so we must calculate residue of dividing †and write it in second mesh of first row (from the left side) beside the module.

Before explaining the table, we remind this point that for solving every congruence equation, at first it is necessary to examine the condition. In congruence equation, if, so equation has answer when "c" is divisible by.

Therefore, for examining the condition, at first always we calculate successive division in below form and write the obtained residue in first row of the table respectively. If the first row of the table is ended to number "1", it shows that condition is established and if it is ended to zero and number before zero is greater than "1", that number is the greatest common divisor of "a" and "n".

Stage (1). We calculate successive divisions in below form and write obtained residues in first row of the table.

 

 

 

1999

1378

1378

621

1378

1

 

1242

2

Oval: 621

 

 

Oval: 136

 

 

 

 

 

 

621

136

3

2

544

4

, Ö ,

2

1

Oval: 77

 

 

Oval: 1

 

 

Stage (2). We write congruencyís constant number that is "1" and "Ė1" successively in the second row of the table (row) from left to right.

Stage (3). We write the first residue (621) to the last residue (1) another time in the third row (row) like the first row (ni).

Stage (4). Always we suppose the last mesh of forth row in right side of table (row) is equal to constant number (+1 or Ė1) of its column in second row. It is necessary to explain that in first row, with condition, the number "1" appears and so, division is finished.

For completing the forth row (row) of the right side in successive "M" figures (according to figure and in direction of arrows)

We act in this method (action of every row, is written at beginning of every row):

We multiply the last number from the right side in forth row (537) in congruence constant number:

In the other hand, obtained residue of dividing number "537(1357)" or "728709" by "1378" is answer of congruence equation:

(Main answer of equation)

It is obvious that general answer of equation for arbitrary integer value of †is:

(General answer of equation)

21.1.3. Example

Solve below congruence equation:††††††††

Solution. Since "", at first we write obtained residue from division in the right side of "module 19". Then we calculate successive divisions and write obtained residues on first row of the table:

According to the above congruence, positive main answer of equation is:

†† ;††††

 

 

19

7

5

2

1

1

-1

1

-1

 

7

5

2

1

 

Oval: -8

-3

-2

-1

 

Oval: -52121.1.4. Example

Find inverse of number "1999" with "module 1980".

Solution. Mathematics expression of problem is below congruence equation:

Since, we write residue of division in the right side of "module 1980" and then we calculate successive divisions.

21.1.5. Attention

†Since always our purpose is positive answer of equation, therefore, it is enough to add number of module (1980) to number (-521) that the main answer of equation is obtained:

1980

19

4

3

1

1

-1

1

-1

 

19

4

3

1

 

 

-5

-2

-1

 

 

So, multiplicative inverse of number "1999" in "module 1980" is "1459".

21.1.6. Example

Verify below congruence equation about existing or non- existing answer:

Solution. At first, we obtain the greatest common divisor of two numbers "3953" and "2680" with successive divisors and writing the residues.

 

3953

2680

1273

134

 

0

1

-1

1

-1

1

 

 

According to the table:

Since constant number of equation (176) isnít divisible by "67", so equation hasnít answer.

 

Oval: 6721.1.7. Example

Solve below congruence equation:

Solution. We solve the equation by forming the table:

()

77

69

8

5

3

2

1

1

-1

1

-1

1

-1

 

69

8

5

3

2

1

 

-29

-26

-3

-2

-1

-1

 

 

One answer of equation from infinite answers is:

Since we want positive answer of equation, therefore, we add positive coefficient of module to this answer:

x=28

 
†††† ;

21.1.8. Example

Find inverse (multiplicative inverse) of number "34" with module "41".

Solution. This problem is equivalent with below congruence:

We solve the equation by forming table:

41

34

7

6

1

1

-1

1

-1

 

34

7

6

1

 

 

-5

-1

-1

 

 

Oval: -6According to the table, answer "" will be obtained that for finding positive answer; it is enough to add the module to it:

x = 35

 
;†††††

Therefore, inverse of number "34" is number "35" with module "41".

21.1.9. Problems

Ÿ By using the table "H.M", solve below congruence equations.

 

1) (

,

2)

3)

,

4) †(: answer)

 

5)

,

6) (: answer)

 

7) ††

,

8)

21.1.10. Attention

This table has a lot of usageís that we explain the usage of table in solving linear Diophantine equations briefly. This table can be used for identifying the prime numbers and finding equivalent of exponential numbers etc.

21.2. Solving linear Diophantine equations by "H.M" table

We consider two-unknown Diophantine equation in below general form:

††††††††††† (1)

We write equation (1) in below congruence equation form:

††††††††††††††††(2)

Now, we solve equation (2) by table and determine value of. Then we obtain value of †from relation.

21.2.1. Example

Solve below Diophantine equation.

††††††† (1)

Solution. At first, we write the equation in below congruence equation form:

†††††††††† (2)

Now, we solve congruence equation (2) by the table:

Oval: -2

17

8

1

1

-1

 

8

1

 

 

-1

 

21.2.2. Attention

†If we have a special answer of equation (1), we can write general answer of

 

equation, because if is a special answer of equation, then:

If, by subtracting the two equations of system:

In order to, it is necessary to have:

Therefore, if, answers of equation (1) will be obtained from below equations:

According to equations (3), general answer of equation †is:

21.2.3. Example

Solve below Diophantine equation.

Solution. At first, we write equation in below congruence equation form:

††††††††††† (1)

Now, we must obtain product of (1340, 3953):

 

3953

1340

1273

 

0

1

-1

1

-1

 

 

Since the first row is ended to "Zero", it is results that numbers "1340" and "3953" have common factor "67":

Oval: 67In the other hand since, therefore we can simplify equation (1) to 67:

Here, for solving equation (2), we form the table:

 

59

20

19

1

1

-1

1

 

20

19

1

 

 

1

1

 

Oval: 3According to the table:

If we have value of, we can calculate value of:

Therefore, the general answer of equation is:

21.2.4. Example

Obtain positive and integer answers of below Diophantine equation:

†††††††† (1)

Oval: -3Solution. The greatest unknown coefficient is coefficient of, therefore:

So, we can write:

††††††††††††††† (2)

With forming the table:

 

11

7

4

3

1

1

-1

1

-1

 

7

4

3

1

 

 

-2

-1

-1

 

 

According to below congruence:

Values of †and †are:

As the same method:

†††††††††† (3)

In this congruence equation, coefficient of unknown is "7" and the module is †according to congruence (2). Therefore, answer of equation (3) according to the recent table is:

Equation (1) hasnít integer answer for †and, so equation has two kinds of special answers †and that we can obtain general answers of equation by using these special answers.

21.2.5. Problems using the table "H.M" solve below Diophantine equations

1) If, find integer numbers †and †so that below equation is established.

2)

3)

4)

5)


21.3. Solving "n-th" degree Diophantine equation by "H.M" table

We consider below "n-th" degree Diophantine equation, we determine positive values of† †and †for it:

†††††††† (1)

For solving equation (1), we write it in below congruence equation form:

†††††††††† (2)

It is obvious that condition of existing answer for equation (2) or (1) is that, so, equation (2) or (1) has "m" two by two non-congruent answers (different answers) with module.

In the other hand, if, we write equation (1) in below form:

††††††††††† (3)

By using the table (H.M), we solve equations (2) or (3).

By using the table (H.M), answers †or will be obtained for (2) and (3) respectively therefore, if, we gain these answers:

††††† (I)†††††† ,†† ††††††††††(II)

Here, If value of† †exists, it is resulted from relation (II):

†††††††††† (4)

If we consider and limits of† †will be determined:

††††††††††††† (5)

Positive and integer values of †are determined according to (5) and relation (II).

For example, we solve below fifth degree equation and we find its positive answers:

††††††††† †(1)

 

 

At first, we write equation (1) in below congruence equation form:

†††††††††††††††† (2)

We simplify equation (2) and then solve it by table (H.M):

†††††††††† (3)

According to the table (H.M):

†††††††††††††††††††††††

29

4

1

1

-1

 

4

1

 

Oval: -7

-1

 

 

Therefore, answers of equation (3) are:

††† †††††† (4)

By substituting the values of† †in (1):

†† †††††† (5)

According to †† "" and "", Suitable value of †is determined from equation (5):

Here, by substituting †in equation (4), †will be obtained:

For another example, we solve below equation and obtain its positive answers:

Since we want positive answers of equation, therefore, we determine limits of †and .

If "" ;† then or †that it is enough to test the

 

main equation only for five primitive natural numbers:

Therefore, an answer of equation is "" and "".

Substituting another four natural numbers instead of †(in main equation), no other positive answer will be obtained.

21.3.1. Remark

If "" and "", then equation (1) is solvable easily, because, we can calculate limits of †and †in this case, it means that we write equation (1) in "" or "" form (""and "y" are positive):

21.3.2. Example

We solve below equation:

Since we want positive answers of above equation, we can calculate limits of† †and †from the main equation:

†††††††††††††††††††††††

By substituting three numbers that have been obtained for, value of †and then †will be obtained. Therefore, we have determined the only positive answer of equation by limits check method. In this method, it is enough to determine the limits of:

21.3.3. Remark

†If †be a prime number, equation (1) changes to below easy and solvable form:

††††††††††††††† (6)

We write equation (6) by using the Fermatís (smaller) theorem †in below form:

††††† ††††† (7)

Answer of equation (7) can be calculated by using the table (H.M) easily.

21.3.4. Example

We solve below 7-th degree Diophantine equation:

†††††††††† (1)

Solution. According to, we divide equation (1) by "2":

†††††††††† (2)

We write equation (2) in below congruence equation form:

†††††† †††††††† †(3)

According to Fermatís (smaller) theorem †equation (3) changes to below equation:

Therefore, answers of equation (1) are in below form:

Here, it is enough to calculate value of† †with respect to:

Answers of equation:

†††††††††††

k

0

1

2

Ö

x

2

9

16

Ö

y

5

683263

651914373

Ö

21.4. A new and fast method for calculating of "" †††††††††††††††††determinant ("H.M" method)

For showing new method of calculation of determinant, it is enough to calculate some determinant in general and numeral cases. Therefore, we calculate third and forth order determinant in general and numeral case.

21.4.1 Calculation of third order determinant

Calculation method. According to the determinant rules always we can suppose that medium element of first column namely †is non-zero. Therefore, we reduce †with chain second order determinant method.

Action method. At first, we write second row another time. In next stage, we put the obtained first column between second and third column another time. Then we separate the obtained element to "four-classes" in a †determinant that we have a †determinant so that one of its elements is a †determinant itself. At last, for calculation of, it is enough to divide the obtained value by.

Action stages. We write the second row another time and we put obtained first column between second and third column another time. Then separate the obtained element to "four-classes":

 

Stage 1:

 

Stage 2:

Every one of "four-classes" forms a †determinant. Then by calculation of every one of †determinants and general determinant and dividing obtained number by, value of †will be calculated.

 

Stage 3:

Therefore †is:

After necessary abridgment:

21.4.2. Example

Calculate below determinant:

 

ŸCalculation with second order chain determinant method

 

 

1) We write the second row another time:

2) We write obtained first column another time:

 

3) We have separated to second order determinants:

4) We calculate the second order chain determinant:

5) We divide the second order chain determinant product by middle element of determinantís first column:

6) Calculated value of:

21.4.3. Simplified method of chain determinant

†† Here, we present simplified and brief method of second order chain determinant. At first, we separate first column and perform actions on successive rows

 

in below form:

(According to the determinants rule, we suppose that "")

Here, we calculate example (1) (chain determinant of second order) by simplified method.

At first we separate the first column and perform actions on rows and then we calculate "".

†††††† ††

2) Calculation of forth order determinant in separation method (chain second order determinant):

With the same method, at first we reduce †to †and then reduce †to chain second order determinant like before.

For separation "" to second order determinants, it is enough to write second and third rows another time and write obtained first column alternately repeatedly and then we perform the separation. Here, according to determinants rules, we suppose , †(like with changing two rows or two columns). Therefore, always we can suppose that ",", then with this hypothesis:

According to the determinant rules, we can suppose that "" and so:

 

Now, we present simplified method of chain second order determinant:

At first we separate first column and then we perform the actions on rows in below form (we suppose):

 

(Final result)

21.4.4. Explain

It is necessary to explain that for calculation of the "n-th" order determinant when, simplified method of chain second order determinant can be used also, because for higher order of determinant, we need a method with higher speed. For calculation, we must attend that first column middle elements of every stage, namely,, †and Ö must be non-zero. For example for calculation , we must have ; because in stage(1) of , we must have , and in stage (2) of , we must have †and in stage (3) of , we must have . It is clear that according to determinants rules for every stage, this thesis is certain and it hasnít any disorder to generality of problem.

21.3.5. Example

Calculate "forth order" determinant by chain second order determinant method.

Calculation with chain second order determinant method:

††††††††††† †

 

Here, we calculate example (21.4.5) by simplified method of chain second order determinant.

In order to this, at first, we separate the first column and perform the calculation on rows in below form (like third order determinant):

 

†††††††††††

 

†††††††††††

†††††††††††

 

21.4.6. Example

Calculate below "fifth order" determinant.

 

 

Solution. We perform calculation on rows and then we obtain, †and †determinant.

 

Product of determinant is:

 

†††††††††††

 

21.5. Definition of regular and irregular prime numbers††††††††††††††††††††††† by "H.M" determinant

†Below Congruence is result of theories related to verification of Fermatís last

 

 

theorem:

Here, Congruence is equivalent with congruence

and below theorems are resulted from congruence (1).

21.5.1. Theorem

Number †is irregular if and only if a number exists for †so that is divisible by.

21.5.2. Theorem

Number †is regular if and only if for every, number

isnít divisible by.

It is easier to consider the general sum:

†††††††††††††††††††††††††††††††††††

It can be proved easily with mathematic induction that numbers, values of polynomial †of †degree for that has rational coefficients (: Bernolliís numbers):

It is interesting that Bernolliís numbers entered into mathematics during solving one of astronomy problems.

The subject is that infinite sequence of Bernolliís numbers appears during converting some simple functions as power series. Like, we can prove:

This series was put against Bernolli during astronomy examinations.

This point that in series related to extension of , only Bernolliís numbers appear that have an even index, isnít randomly, because we can prove easily that all of Bernolliís numbers that have odd index (except number ), are equal to "zero":

If †then

Bernolliís numbers with even index, have interesting characteristics. Like values of famous "Riemann's function" (function), can be expressed with respect to Bernolliís numbers when have even power:

The first Bernolliís numbers are in this form.

†††††††††††††††††††††††

The numerators of these fraction increase rapidly, For example: ††††††††††††††††††††

Suppose:

is indicator of number. Without special difficulty we can prove that for, denominator from number †isnít divisible by.

But, for every, we have this equality:

 

It means:

 

Therefore, for, the number in right side of parenthesis is an integer number.

We return to previous congruence (module) and we simplify it to, so, the result is:

Therefore, congruence "" is established if and only if we have:

Therefore, we can present Kummer's theorem.

21.5.3. Kummerís theorem

Prime number is regular if and only if it is not a divisor of numerators of Bernolliís numbers in "" form.

We express some examples (see the Bernolliís numbers that we expressed before):

isnít divisible by "5".

and †arenít divisible by "7".

and arenít divisible by "11".

†and arenít divisible by "13".

and arenít divisible by "17".

†and arenít divisible by "19".

†and †arenít divisible by "23".

Therefore, all of these powers are regular.

In other word, Fermatís last theorem is correct for powers "5,7,11,13,17,19†††† and 23".

21.4.4. Attention

In next section, you will be acquainted with "H.M" new method of calculation of †that is expressed by a determinant.

We present calculation †by determinant without proof ("H.M" method):

21.5.5. Result

Since †is a prime number and, so.

Number is a regular prim number if and only if †is not divisible by †for every.

In fact, according to identification function of prime numbers, we can write:

 

Therefore, set of regular prime numbers is defined in below form:

(Set of regular prime numbers)

It is obvious that according to definition of set of regular prime numbers, set of irregular prime numbers can be defined by definition of prime numbers set:

(: Set of irregular prime numbers)

†††††††††††††††††††††††

Here, sets of regular and irregular prime numbers are defined by expressing "" and definition of prime numbers set "IP".

 

†Calculation of "". New method for calculating the sum of "k-th" power of †natural numbers †by determinant

Here, we prove that "" calculates by below "k-th" order determinant:

 

†††††††††††

21.5.8. Result

Fermatís last theorem is correct for powers.†

21.5.9. Important result

Other sets can be defined by prime numbers formula.

 

Here, sets of regular and irregular prime numbers are defined by expressing †and definition of prime numbers set "".

21.6. New method of calculation of sum of "k-th" power of the first "n" natural numbers by "H.M" determinant††††††††††††††††††††††††††††††††††† (expressing "" by a determinant)

Here, we prove that "" can be calculated by below "k-th" order determinant:

†††††††††††††††††††††††

 

Proof: We know that if †then:

We put, and †in above identity:

†††††††††††

Now, if we put values †respectively in above identity, we have below equality with adding obtained relations:

We write above equality in below form:

Now, if we put values †respectively for, below system will be obtained:

Text Box: (Trigonometric) system)

So:

 

21.7. Determining the number of roots of perfect cubic degree†††† equation directly by "H.M" method

†††††††† ††† ††††††††(1)

In equation (1), if we have:

then equation has a real root that is computable easily, therefore for calculating the root of equation (1) it is enough to perform this conversion †and establish the above mentioned conditions. As we know, after conversion we will have below equation:

††††† (2)

Now, for calculating, we establish the first condition namely. After establishing this condition we will see that value of †is not obtained.

Now, we establish the second condition, namely. And we will see that value of †will be calculated. Therefore:

††† (3)

After summarizing relation (3), we have below second degree equation (with respect to):

†††††††††† (4)

Here, for calculating the value of, it is enough to solve the second degree equation (4). After solving equation (4), the value of †is:

†††††††† (5)

Discriminant †of cubic degree equation (1) is obtained from value of:

†††††††††† (6)

Here, with obtaining such equation, we can discuss about the number of its roots:

1) If, then equation (1) has a real root and two complex roots.

2) If, then equation (1) has an additive root and one simple root.

3) If, then equation (1) has three real roots.

21.7.1. Remark

In equation (1), If, then the equation is:

†††††††††† (7)

It means the focal equation of cubic degree will obtained. Now, if we put value †in Discriminant of equation (1), Discriminant †of focal equation (7) will obtained:

 

 

21.8. Proof of a new and applied "H.M" theorem

(Concerning the factorization of composite numbers)

21.8.1. "H.M" Theorem

If† †is divisible by "p" and result of is an odd number, then:

Proof. †For every †and , we can write:

††††††††††††††††††††††† †††††††††††††† (1)

If †is odd and according to below equality for every odd:

(: Number integer part)

†††††† ( is odd number)††††††††††

So:

†††††††† (2)

21.8.2. Example

†If †is prime number, according to Wilsonís theorem, expression †is divisible by †and conversely (If †is divisible by and †is a prime number). Therefore, if †is factorization in relation (2) form, †is a prime number:

†††††††††† (3)

21.8.3. Example

21.8.4. Example

I. If †then:

(p: prime number)††††† †

II. If then: