## New Articles

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*Ref*]**example of contractive sequence**by pahioSep 20[

*Ref*]**contractive sequence**by pahioAug 29[

*Ref*]**Some formulas of partnership**by burgessAug 26[

*Rec*]**Kenosymplirostic numbers**by imaginary.iAug 14[

*Edu*]**How to find whether a given number is prime or not...**by burgessAug 12[

*Edu*]**BODMAS Rule application**by burgessAug 8[

*Edu*]**Tests of Divisibility- Simple tricks**by burgessAug 7[

*Res*]**0/0 is possible and has an answer**by imaginary.iAug 2[

*Ref*]**Sophomore's dream**by pahioJul 9[

*Res*]**examples of growth of perturbations in chemical or...**by rspuzioMay 24[

*Ref*]**proof of Stirling's approximation**by rspuzioMay 8[

*Res*]**Example of stochastic matrix of mapping**by rspuzioApr 23[

*Res*]**6. Discussion**by rspuzioApr 20[

*Res*]**5. Entanglement**by rspuzioApr 20## Latest Messages

Sep 23

Sep 23

Sep 19

Sep 17

Sep 15

Sep 14

Sep 14

Sep 12

Sep 10

Sep 9

Sep 8

Sep 3

Sep 1

Aug 31

hi bro, i think this tip will help you :
- Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
and i found an harshad number digit sum 19:
874

Sep 23

We can construct pseudoprimes in k(i) as follows:
Take a couple of primes in k(1) of the same type i.e. both
being of type 4m+3 or both being of type 4m+1.
Let us take 3*7 = 21 as an example of first type. Choose
as base 21 + i. ((21+i)^20-1)/21 yields a quotient which is
a Gaussian integer; hence 21 is a psedoprime in k(i).
Note a) This is possible only with the aid of software pari
or similar. b) 21 + i, 21 - i, 1 - 21i, -1+21i, all yield
identical Gaussian integers when we apply Fermat's theorem.
b) Second base is chosen as follows: Partition 22 into two parts
such that one is divisible by 3 and the other by 7 i.e.
7 + 15i, 7 - 15i, -7 + 15i and -7 - 15i are four different
points on the complex plane which yield identical Gaussian integers
when Fermat's theorem is applied. Similar remarks apply
to composites in k(1) each of which is of type 4m + 1.

Sep 19

When we apply Fermat's theorem to four different points in the
complex plane we get an invariant result; the four different points
are 21 + i, -21 - i, 1 - 21i and -1 + 21i. i.e. ((21 + i)^20-1)/21=
((-21-i)^20-1)/21 = ((1-21i)^20)/21 = ((-1+21i)^20/21.

Sep 17

Although the above is not functioning I am able to read my older
messages by clicking on "older". However wish the search facility
is restored early.

Sep 15

Regretable that search facility is still not functioning.

Sep 14

Hi parag,
I think you will find the answer to your question in the PM entry Heron's principle (http://planetmath.org/heronsprinciple). Unfortunately, the entry is now almost impossible to find because the PM SEARCH ENGINE DOES NOT FUNCTION :=(
I'm also sorry that the image for the proof IS NOT VISIBLE, AS NOT ARE NOW ALL IMAGES IN PLANETMATH.
Regards,
Jussi

Sep 14

Primes of shape p'= 4m + 3 are prime in both k(1) and k(i). With
reference to Frermat's theorem in k(i) what is the nature of
the bases? In fact they form a group isomorphic with z_n; the
bases are given by (1 + kp' + i). Let me illustrate this with
examples p' = 11 and 19. Here k belongs to W.
The congruence in all these cases is to -i and not 1; recall that
-i is also one of the unities in k(i). Hence ((1 + i)^10 + i)/11
= 3i and ((12 + i)^10 + i)/11 = 3926445133 + 4305498635i.
((1 + i)^18 + i)/19 and ((20 + i)^18 + i)/19 are also equal to
Gaussian integers.

Sep 12

341 is a pseudoprime in k(1). It is also a pseudoprime in k(i) i.e.
((1+i)^340 + 1)/341 = a Gaussian integer. Note it is +1 in parenthesis
because -1 is also a unity in k(i). Also 1105, a Carmichael number
in k(1) is a pseudoprime in k(i) i.e. ((6 + i)^1104 -1)/1105 =
a Gaussian integer . However, it is too big to be copied here.

Sep 10

I have posted many messages; however I do not know how many have
viewed the messages. Request administrators to instal software
to enable this.

Sep 9

Some more examples pertaining to the previous message:
a) ((6+i)^4-1)/5 = 216 + 168i b)((6 + i)^12-1)/13 = - 78849720
+ 180928440i c) ((2 +3i)^4-1)/5 = -24 +24i

Sep 8

Any prime in k(1) with shape 4m+1 can be factorised in k(i).
Examples : 5 = (2+i)(2-i), 13 = (2+3i)(2-3i) or (3 +2i)(3 - 2i).
Now if we use any one of such factors and apply Fermat's theorem
with respect to a prime of the same shape ( excepting the prime
in k(1) of which the base is a factor) we get quotients which
are integers in k(i). Examples: a)((2+i)^12-1)/13 = 904 - 782i
b) ((2+i)^16-1)/17 = 9696 + 20832i. However neither 13 nor 17
are primes in k(i) - in fact they are pseudo primes in k(i).

Sep 3

In one of my recent messages I had stated that there are
four unities in k(i) viz 1, -1, i and -i. Fermat's theorem
holds true in k(i) when we keep this in mind. For example
(2+I) and 3 are co-prime. Hence ((2+I)^4 + 1)/3 = -2 + 8i. I
will be giving a few more examples in the next message.

Sep 1

Numbers of the type 4m+1 are not prime in k(i). However their
factors are prime. Example: 5 = (2+i)(2-i). If we take one of these
as the base Fermat's theorem works with respect to other
co-primes of the type 4m+1. Examples:
a) ((2-i)^12-1)/13 = 904 - 792i b)((2+i)^16-1)/17 = 9696 + 20832i

Aug 31

Thanks for the references. Thomas Edison scaled down the building blocks of electricity to economically feasible units. Edison's first scaled model was 1/3 copper, cost-wise. The scaling down of an economic system to finite units was first implemented by Middle Kingdom Egyptians, a finite unit fraction methodology that has not been fully decoded. Greeks used the finite model. Arabs and medieval scribes modified the multiplication scaled model to a subtraction model, a system that ended with Galileo's inverse proportion square root method.

## Latest Messages

Sep 23

Sep 23

Sep 19

Sep 17

Sep 15

Sep 14

Sep 14

Sep 12

Sep 10

Sep 9

Sep 8

Sep 3

Sep 1

Aug 31

hi bro, i think this tip will help you :
- Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
and i found an harshad number digit sum 19:
874

Sep 23

We can construct pseudoprimes in k(i) as follows:
Take a couple of primes in k(1) of the same type i.e. both
being of type 4m+3 or both being of type 4m+1.
Let us take 3*7 = 21 as an example of first type. Choose
as base 21 + i. ((21+i)^20-1)/21 yields a quotient which is
a Gaussian integer; hence 21 is a psedoprime in k(i).
Note a) This is possible only with the aid of software pari
or similar. b) 21 + i, 21 - i, 1 - 21i, -1+21i, all yield
identical Gaussian integers when we apply Fermat's theorem.
b) Second base is chosen as follows: Partition 22 into two parts
such that one is divisible by 3 and the other by 7 i.e.
7 + 15i, 7 - 15i, -7 + 15i and -7 - 15i are four different
points on the complex plane which yield identical Gaussian integers
when Fermat's theorem is applied. Similar remarks apply
to composites in k(1) each of which is of type 4m + 1.

Sep 19

When we apply Fermat's theorem to four different points in the
complex plane we get an invariant result; the four different points
are 21 + i, -21 - i, 1 - 21i and -1 + 21i. i.e. ((21 + i)^20-1)/21=
((-21-i)^20-1)/21 = ((1-21i)^20)/21 = ((-1+21i)^20/21.

Sep 17

Although the above is not functioning I am able to read my older
messages by clicking on "older". However wish the search facility
is restored early.

Sep 15

Regretable that search facility is still not functioning.

Sep 14

Hi parag,
I think you will find the answer to your question in the PM entry Heron's principle (http://planetmath.org/heronsprinciple). Unfortunately, the entry is now almost impossible to find because the PM SEARCH ENGINE DOES NOT FUNCTION :=(
I'm also sorry that the image for the proof IS NOT VISIBLE, AS NOT ARE NOW ALL IMAGES IN PLANETMATH.
Regards,
Jussi

Sep 14

Primes of shape p'= 4m + 3 are prime in both k(1) and k(i). With
reference to Frermat's theorem in k(i) what is the nature of
the bases? In fact they form a group isomorphic with z_n; the
bases are given by (1 + kp' + i). Let me illustrate this with
examples p' = 11 and 19. Here k belongs to W.
The congruence in all these cases is to -i and not 1; recall that
-i is also one of the unities in k(i). Hence ((1 + i)^10 + i)/11
= 3i and ((12 + i)^10 + i)/11 = 3926445133 + 4305498635i.
((1 + i)^18 + i)/19 and ((20 + i)^18 + i)/19 are also equal to
Gaussian integers.

Sep 12

341 is a pseudoprime in k(1). It is also a pseudoprime in k(i) i.e.
((1+i)^340 + 1)/341 = a Gaussian integer. Note it is +1 in parenthesis
because -1 is also a unity in k(i). Also 1105, a Carmichael number
in k(1) is a pseudoprime in k(i) i.e. ((6 + i)^1104 -1)/1105 =
a Gaussian integer . However, it is too big to be copied here.

Sep 10

I have posted many messages; however I do not know how many have
viewed the messages. Request administrators to instal software
to enable this.

Sep 9

Some more examples pertaining to the previous message:
a) ((6+i)^4-1)/5 = 216 + 168i b)((6 + i)^12-1)/13 = - 78849720
+ 180928440i c) ((2 +3i)^4-1)/5 = -24 +24i

Sep 8

Any prime in k(1) with shape 4m+1 can be factorised in k(i).
Examples : 5 = (2+i)(2-i), 13 = (2+3i)(2-3i) or (3 +2i)(3 - 2i).
Now if we use any one of such factors and apply Fermat's theorem
with respect to a prime of the same shape ( excepting the prime
in k(1) of which the base is a factor) we get quotients which
are integers in k(i). Examples: a)((2+i)^12-1)/13 = 904 - 782i
b) ((2+i)^16-1)/17 = 9696 + 20832i. However neither 13 nor 17
are primes in k(i) - in fact they are pseudo primes in k(i).

Sep 3

In one of my recent messages I had stated that there are
four unities in k(i) viz 1, -1, i and -i. Fermat's theorem
holds true in k(i) when we keep this in mind. For example
(2+I) and 3 are co-prime. Hence ((2+I)^4 + 1)/3 = -2 + 8i. I
will be giving a few more examples in the next message.

Sep 1

Numbers of the type 4m+1 are not prime in k(i). However their
factors are prime. Example: 5 = (2+i)(2-i). If we take one of these
as the base Fermat's theorem works with respect to other
co-primes of the type 4m+1. Examples:
a) ((2-i)^12-1)/13 = 904 - 792i b)((2+i)^16-1)/17 = 9696 + 20832i

Aug 31

Thanks for the references. Thomas Edison scaled down the building blocks of electricity to economically feasible units. Edison's first scaled model was 1/3 copper, cost-wise. The scaling down of an economic system to finite units was first implemented by Middle Kingdom Egyptians, a finite unit fraction methodology that has not been fully decoded. Greeks used the finite model. Arabs and medieval scribes modified the multiplication scaled model to a subtraction model, a system that ended with Galileo's inverse proportion square root method.