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 "kth" 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)
_{} (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.
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 

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 













621 
136 

3 
2 

544 
4 
, … , 
2 
1 













Stage (2). We write congruency’s constant number that is "1" and "–1" 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 (n_{i}).
Stage (4). Always we suppose the last mesh of forth row in right side of table (row_{}) is equal to constant number (+1 or –1) 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) _{}
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 

_{} 

3 
2 
1 

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.
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".
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) isn’t divisible by "67", so equation hasn’t answer.
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:

_{} ;
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 

According
to the table, answer "_{}" will be obtained that for
finding positive answer; it is enough to add the module to it:

_{};
Therefore, inverse of number "34" is number "35" with module "41".
Ÿ By using the table "H.M", solve below congruence equations.
1) _{}(_{} 
, 
2) _{} 
3) _{} 
, 
4) _{} (_{}: answer)

5) _{} 
, 
6) _{}(_{}: answer) 
7) _{} 
, 
8) _{} 
This table has a lot of usage’s 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.
_{}
We consider twounknown 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_{}.
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:
_{}
_{} 
17 
8 
1 
_{} 
1 
1 

_{} 
8 
1 

_{} 

1 

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:
_{}
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":
_{}
In 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 

According to the table:
_{}
If we have value of_{}, we can calculate value of_{}:
_{}
Therefore, the general answer of equation is:
_{}
Obtain positive and integer answers of below Diophantine equation:
_{}
(1)
Solution. 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) hasn’t 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.
1) If_{}, find integer numbers _{} and _{} so that below equation is established.
_{}
2) _{}
3) _{}
4) _{}
5) _{}
We consider below "nth" 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 noncongruent 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 

_{} 

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.
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):
_{}
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_{}:
_{}
If _{} be a prime number, equation (1) changes to below easy and solvable form:
_{} (6)
We write equation (6) by using the Fermat’s (smaller) theorem _{} in below form:
_{} (7)
Answer of equation (7) can be calculated by using the table (H.M) easily.
We solve below 7th 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 Fermat’s (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 
… 
x 
2 
9 
16 
… 
y 
5 
683263 
651914373 
… 
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.
_{}
Calculation method. According to the determinant rules always we can suppose that medium element of first column namely _{} is nonzero_{}. 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 "fourclasses" 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 "fourclasses":
Stage 1:
_{}
Stage 2:
_{}
Every one of "fourclasses" 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:
_{}
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 determinant’s first column:
_{}
6) Calculated value of_{}:
_{}
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)
It is necessary to explain that for calculation of the "nth" 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 … must be nonzero. 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 hasn’t any disorder to generality of problem.
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):
_{}
_{}
_{}
Calculate below "fifth order" determinant.
_{}
Solution. We perform calculation on rows and then we
obtain_{}, _{} and _{} determinant.
_{}
Product of determinant is:
_{}
Below Congruence is result of theories related to verification of Fermat’s last
theorem:
_{}
Here, Congruence _{}is equivalent with congruence
_{}and below theorems are resulted from congruence (1).
Number _{} is irregular if and only if a number exists for _{} so that _{}is divisible by_{}.
Number _{} is regular if and only if for every_{}, number
_{}
isn’t 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 (_{}: Bernolli’s numbers):
_{}
It is interesting that Bernolli’s numbers entered into mathematics during solving one of astronomy problems.
_{}
The subject is that infinite sequence of Bernolli’s 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 Bernolli’s numbers appear that have an even index, isn’t randomly, because we can prove easily that all of Bernolli’s numbers that have odd index (except number _{}), are equal to "zero":
If _{} then _{}
Bernolli’s numbers with even index_{}, have interesting characteristics. Like values of famous "Riemann's function" (function_{}), can be expressed with respect to Bernolli’s numbers when have even power:
_{}
The first Bernolli’s 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 _{} isn’t 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.
Prime number _{}is regular if and only if it is not a divisor of numerators of Bernolli’s numbers in "_{}" form.
We express some examples (see the Bernolli’s numbers that we expressed before):
_{}isn’t divisible by "5".
_{}and _{} aren’t divisible by "7".
_{}and _{}aren’t divisible by "11".
_{} and _{}aren’t divisible by "13".
_{}and _{}aren’t divisible by "17".
_{} and _{}aren’t divisible by "19".
_{} and _{} aren’t divisible by "23".
Therefore, all of these powers are regular.
In other word, Fermat’s last theorem is correct for powers "5,7,11,13,17,19 and 23".
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):
_{}
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 "kth" power of _{} natural numbers _{} by determinant
Here, we prove that "_{}" calculates by below "kth" order determinant:
_{}
Fermat’s last theorem is correct for powers_{}.
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 "_{}".
_{}
Here, we prove that "_{}" can be calculated by below "kth" 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_{}:
_{}
So:
_{}
_{} (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.
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:
_{}
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)
If _{} is prime number, according to Wilson’s 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)
I. If _{} then:
(p: prime number) _{}
II. If _{}then: