Fork me on GitHub

planetmath.org

Math for the people, by the people.

Welcome!

PlanetMath is a virtual community which aims to help make mathematical knowledge more accessible. PlanetMath's content is created collaboratively: the main feature is the mathematics encyclopedia with entries written and reviewed by members. The entries are contributed under the terms of the Creative Commons By/Share-Alike License in order to preserve the rights of authors, readers and other content creators in a sensible way. We use LaTeX, the lingua franca of the worldwide mathematical community. On February 13th 2013, PlanetMath.org was updated to use the new software system Planetary. Some release notes are here. Please report bugs in the Planetary Bugs Forum or on Github.

User login



Latest Messages  

[p] my help for you by mohamed benabbou 1:09 pm
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

[p] Constructing pseudoprimes in k(i) by akdevaraj 12:31 pm
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.

[p] pseudoprimes in k(i) - invariants by akdevaraj 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.

[p] Search facility by akdevaraj 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.

[p] Search facility by akdevaraj Sep 15
Regretable that search facility is still not functioning.

[p] Heron's principle (geometry) by pahio 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

[P] Fermat's theorem in k(i) - base groups by akdevaraj 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.

[P] pseudoprimes in k(i) (contd) by akdevaraj 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.

[P] viewership of messages - II by akdevaraj 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.

[P] pseudoprimes in k(i) (contd) by akdevaraj 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

[P] pseudoprimes in k(i) (contd) by akdevaraj 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).

[P] Fermat's theorem in k(i) (c0ntd) by akdevaraj 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.

[P] Fermat's theorem in k(i) (c0ntd) by akdevaraj 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

[P] Enlanglement as a basic modeling idea by milogardner 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.