## New Articles

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*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[

*Res*]**4. Measurement**by rspuzioApr 20## Latest Messages

Sep 14

Sep 14

Sep 12

Sep 10

Sep 9

Sep 8

Sep 3

Sep 1

Aug 31

Aug 31

Aug 30

Aug 28

Aug 25

Aug 22

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.

Aug 31

On the lines of definition of co-primality in k(1) (ref: An
introduction to the theory of numbers by Hardy and Wright )
co-primality in k(i) is defined as follows:
Let a and b be two Gaussian integers in k(i). Then if (a,b )
= 1 or i then a and b are co-prime.

Aug 30

I have written many messages (posts ). However I do not know
how many read my messages. Perhaps rspuzio will consider
setting up software to enable viewership.

Aug 28

When we carry out the following operations we get a quotient
which is a Gaussian integer: a)((21+i)^20-1)/21 b)((21+i)^(21-i)-1)/21
c)((21+i)^(21-i)-i/21 and d)((341+i)^340-1)/341
Note that a) 341 is also a pseudoprime in k(1) and b) i is also one of the unities
in k(i).

Aug 25

Dear planetmath.org:
It has been several months since I have been able to perform a successful search at your site. The fact that I have not been able to find articles either by searching, nor by MSC has rendered (at least for me) planetmath.org practically unusable.
In your home page, I see that new articles are being contributed regularly. This makes me wonder: Is there a way to sidestep the search problem? Is there am alternative way to get to an article (say, "Pascal's formula", or "Pascal's rule")? Is there an alternative way of reaching articles under MSC 05A10?
Thanks in advance.>

Aug 22

In 1987, Egyptologist Gay Robins, and Charles Shute, wrote a book on the Rhind Mathematical Papyrus (RMP). Five years later Egyptologist John Legon wrote on the KP and the same class of arithmetic proportions used in the RMP. The KP and RMP report the same arithmetic proportion method to find the largest term. The method: take 1/2 of the difference, 1/2 of 5/6 (5/12 in the KP) times the number of differences (nine times 5/12 = 15/4 in the KP) plus the sum of the A.P progression (100 in the KP) divided by the number of terms (10 meant 100/10 = 10 in the KP). Finally add column 11’s result, 3 3/4, to 10, and the largest term, 13 3/4.
In unit fractions, the context of the text, add column 11: 5/12 times 9 writing 3 3/4 as 3 2/3 1/12 to 10 in column 12 beginning with the largest term 13 2/3 1/12. The scribe subtracted 5/6 nine times created remaining terms of the arithmetic progression.
Robins-Shute confused aspects of the problem by omitting the sum divided by the number of terms, a topic cited in a closely related RMP 40 problem. A scribal algebraic statement matched pairs added to 20 reporting five pairs summed to 100, a set of facts included in RMP 40.
The complete KP method found the largest term facts reported in RMP 64 and RMP 40 by John Legon in 1992. Scholars have parsed Rhind Mathematical Papyrus 40 a problem that asked that 100 loaves of bread to be shared between five men by finding the smallest term of an arithmetic progression.
C. A confirmation of the Kahun Papyrus arithmetic progression method must include discussions of RMP 40 and RMP 64. In RMP 64 Ahmes asked 10 men to share 10 hekats of barley with a differential of 1/8 defining an arithmetical progression. Robins and Shute reported: ”the scribe knew the rule that, to find the largest term of the arithmetical progression, he must add half the difference to the average number of terms as many times as there are common differences, that is, one less than the number of terms”.
1. number of terms: 10
2. arithmetical progression difference: 1/8
3. arithmetic progression sum: 10
The scribe used the following facts to find the largest term.
1. one-half of differences, 1/16, times number of terms minus one, 9,
1/16 times 9 = 9/16
2. The computed parameter(1), was found by 10, the sum, divided by 10, the number of terms. It was inserted by Robins-Shute, but had not been high-lighted, citing 1 + 1/2 + 1/16, or 1 9/16, the largest term. The remaining nine terms were found by subtracting 1/8 nine times to obtain the remaining barley shares.
That is, the KP scribe used formula 1.0:1111111
(1/2)d(n-1) + S/n = Xn (formula 1.0)
with,
d = differential, n = number of terms in the series, S = sum of the series, Xn = largest term in the series allowed three(of the four) parameters: d, n, S and Xn, to algebraically find the fourth parameter.
When n was odd, x (n/2) = S/n,
and x 1 + xn = x2 + x(n -1) = x3 + x(n -2) = … = x(n/2) = S/n,
Note that Robins-Shute omitted the sum divided by the number of terms (S/n):
A modern footnote cites Carl Friedrich Gauss implementing as a grammar school student a solution to the n = even case. Ahmes and Gauss found the sum for 1 to 100 by using d = 1 following the same rule. Ahmes and Gauss reached the sum 5050 based on 50 pairs of 101 (1 + 101 = 2 + 99 = 3 + 98 = …) by using an identical arithmetic progression rule.

## Latest Messages

Sep 14

Sep 14

Sep 12

Sep 10

Sep 9

Sep 8

Sep 3

Sep 1

Aug 31

Aug 31

Aug 30

Aug 28

Aug 25

Aug 22

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.

Aug 31

On the lines of definition of co-primality in k(1) (ref: An
introduction to the theory of numbers by Hardy and Wright )
co-primality in k(i) is defined as follows:
Let a and b be two Gaussian integers in k(i). Then if (a,b )
= 1 or i then a and b are co-prime.

Aug 30

I have written many messages (posts ). However I do not know
how many read my messages. Perhaps rspuzio will consider
setting up software to enable viewership.

Aug 28

When we carry out the following operations we get a quotient
which is a Gaussian integer: a)((21+i)^20-1)/21 b)((21+i)^(21-i)-1)/21
c)((21+i)^(21-i)-i/21 and d)((341+i)^340-1)/341
Note that a) 341 is also a pseudoprime in k(1) and b) i is also one of the unities
in k(i).

Aug 25

Dear planetmath.org:
It has been several months since I have been able to perform a successful search at your site. The fact that I have not been able to find articles either by searching, nor by MSC has rendered (at least for me) planetmath.org practically unusable.
In your home page, I see that new articles are being contributed regularly. This makes me wonder: Is there a way to sidestep the search problem? Is there am alternative way to get to an article (say, "Pascal's formula", or "Pascal's rule")? Is there an alternative way of reaching articles under MSC 05A10?
Thanks in advance.>

Aug 22

In 1987, Egyptologist Gay Robins, and Charles Shute, wrote a book on the Rhind Mathematical Papyrus (RMP). Five years later Egyptologist John Legon wrote on the KP and the same class of arithmetic proportions used in the RMP. The KP and RMP report the same arithmetic proportion method to find the largest term. The method: take 1/2 of the difference, 1/2 of 5/6 (5/12 in the KP) times the number of differences (nine times 5/12 = 15/4 in the KP) plus the sum of the A.P progression (100 in the KP) divided by the number of terms (10 meant 100/10 = 10 in the KP). Finally add column 11’s result, 3 3/4, to 10, and the largest term, 13 3/4.
In unit fractions, the context of the text, add column 11: 5/12 times 9 writing 3 3/4 as 3 2/3 1/12 to 10 in column 12 beginning with the largest term 13 2/3 1/12. The scribe subtracted 5/6 nine times created remaining terms of the arithmetic progression.
Robins-Shute confused aspects of the problem by omitting the sum divided by the number of terms, a topic cited in a closely related RMP 40 problem. A scribal algebraic statement matched pairs added to 20 reporting five pairs summed to 100, a set of facts included in RMP 40.
The complete KP method found the largest term facts reported in RMP 64 and RMP 40 by John Legon in 1992. Scholars have parsed Rhind Mathematical Papyrus 40 a problem that asked that 100 loaves of bread to be shared between five men by finding the smallest term of an arithmetic progression.
C. A confirmation of the Kahun Papyrus arithmetic progression method must include discussions of RMP 40 and RMP 64. In RMP 64 Ahmes asked 10 men to share 10 hekats of barley with a differential of 1/8 defining an arithmetical progression. Robins and Shute reported: ”the scribe knew the rule that, to find the largest term of the arithmetical progression, he must add half the difference to the average number of terms as many times as there are common differences, that is, one less than the number of terms”.
1. number of terms: 10
2. arithmetical progression difference: 1/8
3. arithmetic progression sum: 10
The scribe used the following facts to find the largest term.
1. one-half of differences, 1/16, times number of terms minus one, 9,
1/16 times 9 = 9/16
2. The computed parameter(1), was found by 10, the sum, divided by 10, the number of terms. It was inserted by Robins-Shute, but had not been high-lighted, citing 1 + 1/2 + 1/16, or 1 9/16, the largest term. The remaining nine terms were found by subtracting 1/8 nine times to obtain the remaining barley shares.
That is, the KP scribe used formula 1.0:1111111
(1/2)d(n-1) + S/n = Xn (formula 1.0)
with,
d = differential, n = number of terms in the series, S = sum of the series, Xn = largest term in the series allowed three(of the four) parameters: d, n, S and Xn, to algebraically find the fourth parameter.
When n was odd, x (n/2) = S/n,
and x 1 + xn = x2 + x(n -1) = x3 + x(n -2) = … = x(n/2) = S/n,
Note that Robins-Shute omitted the sum divided by the number of terms (S/n):
A modern footnote cites Carl Friedrich Gauss implementing as a grammar school student a solution to the n = even case. Ahmes and Gauss found the sum for 1 to 100 by using d = 1 following the same rule. Ahmes and Gauss reached the sum 5050 based on 50 pairs of 101 (1 + 101 = 2 + 99 = 3 + 98 = …) by using an identical arithmetic progression rule.