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*Res*]**examples of growth of perturbations in chemical or...**by rspuzioMay 24## Latest Messages

Sep 30

Sep 28

Sep 28

Sep 28

Sep 28

Sep 23

Sep 23

Sep 19

Sep 17

Sep 15

Sep 14

Sep 14

Sep 12

Sep 10

This is to remind administration about my request to either
a) enable " copy and paste operation" on the templates reserved
for articles and messages or b) open a page on facebook which
will automtically enable copying and pasting as well as uploading
snapshots of articles and messages.

Sep 28

Hi parag,
There is no set theory needed for the proof. I think only the
expressions of m and M are strange for you; they simply mean
that m is the least and M the greatest of the given fractions!
Jussi

Sep 28

I am not able to understand that proof at planetmath.org/summednumeratorandsummeddenominator as I am not aware of the set theory. Please give a simple proof for that.

Sep 28

parag,
do you mean such as in
http://planetmath.org/summednumeratorandsummeddenominator?

Sep 28

Many of the Gaussian integers indicated in "Fermat's theorem
in k(i)" and " pseudoprimes in k(i) " are too big to be copied manually.
I, therefore, suggest that the template for articles and
messages be modified to enable "copy and paste" operations. Alternately
planetmath.org open a page on facebook. This will enable us to
upload snapshots of messages/articles.

Sep 23

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.

## Latest Messages

Sep 30

Sep 28

Sep 28

Sep 28

Sep 28

Sep 23

Sep 23

Sep 19

Sep 17

Sep 15

Sep 14

Sep 14

Sep 12

Sep 10

This is to remind administration about my request to either
a) enable " copy and paste operation" on the templates reserved
for articles and messages or b) open a page on facebook which
will automtically enable copying and pasting as well as uploading
snapshots of articles and messages.

Sep 28

Hi parag,
There is no set theory needed for the proof. I think only the
expressions of m and M are strange for you; they simply mean
that m is the least and M the greatest of the given fractions!
Jussi

Sep 28

I am not able to understand that proof at planetmath.org/summednumeratorandsummeddenominator as I am not aware of the set theory. Please give a simple proof for that.

Sep 28

parag,
do you mean such as in
http://planetmath.org/summednumeratorandsummeddenominator?

Sep 28

Many of the Gaussian integers indicated in "Fermat's theorem
in k(i)" and " pseudoprimes in k(i) " are too big to be copied manually.
I, therefore, suggest that the template for articles and
messages be modified to enable "copy and paste" operations. Alternately
planetmath.org open a page on facebook. This will enable us to
upload snapshots of messages/articles.

Sep 23

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.