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Gaussian Integers and Pythagorean Triplets

by ramsey2879
Tags: gaussian, integers, pythagorean, triplets
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ramsey2879
#1
Sep22-07, 05:52 PM
P: 894
It is well known that 4n(n+1) + 1 is a square if n is an integer but if n is a Gaussian integer i.e., 4n(n+1) + 1 = A + Bi, then the norm (A^2 + B^2) is always a square! The proof is quite easy since A = u^2 - v^2 and B = 2uv.
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robert Ihnot
#2
Sep23-07, 12:41 AM
P: 1,059
4n(n+1)+1 =(2n+1)^2, but it can not equal A+Bi, because you are not considering the imaginary part, are you?

If, instead, we take the form Z=A+Bi, and look at (2Z+1)^2, what do we do now?
ramsey2879
#3
Sep23-07, 02:12 PM
P: 894
Quote Quote by robert Ihnot View Post
4n(n+1)+1 =(2n+1)^2, but it can not equal A+Bi, because you are not considering the imaginary part, are you?

If, instead, we take the form Z=A+Bi, and look at (2Z+1)^2, what do we do now?
If n=Z = 1+2i for example
[tex]4*(1+2i)*(2+2i) + 1 = -7+24i = (3+4i)^2 = (2Z+1)^2 = (2A+1 +2Bi)^2)[/tex]
[tex]-7 = (2A+1)^2 - (2B)^2 = u^2 - v^2[/tex]
[tex] 24 = 2(2A+1)(2B) = 2uv[/tex]
So we have the x and y of the Pathagorean triple: (7*7 + 24*24 = 25*25)

robert Ihnot
#4
Sep23-07, 11:51 PM
P: 1,059
Gaussian Integers and Pythagorean Triplets

First let Z* = conjugate of Z, then if (2Z+1)=A+Bi, we have (2Z*+1)(2Z+1) =A^2 +B^2
This then results in (2a+1)^2 + b^2 = A^2+B^2, which is what is expected. But it does not make (2Z+1)(2Z*+1) a square.

You say, quote: The proof is quite easy since A = u^2 - v^2 and B = 2uv.

You introduce the above, in the second sentence, which is what is required to show that A^2+B^2 =(u^2+v^2)^2. But it does not relate to (2Z+1)^2, as introduced in the first sentence.

Anyway, since with sentence 2 you have created the Pythagorian triples, we need only say for example that since 3^2+4^2 = 5^2, the Gaussian integer 3+4i has a square as its norm.
ramsey2879
#5
Sep24-07, 08:51 AM
P: 894
Quote Quote by robert Ihnot View Post
First let Z* = conjugate of Z, then if (2Z+1)=A+Bi, we have (2Z*+1)(2Z+1) =A^2 +B^2
This then results in (2a+1)^2 + b^2 = A^2+B^2, which is what is expected. But it does not make (2Z+1)(2Z*+1) a square.

You say, quote: The proof is quite easy since A = u^2 - v^2 and B = 2uv.

You introduce the above, in the second sentence, which is what is required to show that A^2+B^2 =(u^2+v^2)^2. But it does not relate to (2Z+1)^2, as introduced in the first sentence.

Anyway, since with sentence 2 you have created the Pythagorian triples, we need only say for example that since 3^2+4^2 = 5^2, the Gaussian integer 3+4i has a square as its norm.
You must multiply 4*Z*(Z+1) and add 1 to get a square that I am talking about.
However any Gaussian integer squared is of the form A+Bi where A= u^2 - v^2 and B = 2uv since I read that the norm of a product of two complex numbers is the product of their norms. So yes it is possible to have two squares that do not sum to a square but that is not possible for the A and B where A+Bi is the square of a Gaussian integer.
ramsey2879
#6
Sep24-07, 09:42 AM
P: 894
Quote Quote by ramsey2879 View Post
You must multiply 4*Z*(Z+1) and add 1 to get a square that I am talking about.
However any Gaussian integer squared is of the form A+Bi where A= u^2 - v^2 and B = 2uv since I read that the norm of a product of two complex numbers is the product of their norms. So yes it is possible to have two squares that do not sum to a square but that is not possible for the A and B where A+Bi is the square of a Gaussian integer.
The last couple of posts are troubling at first glance. Of course the product of two conjugates equals A^2 + B^2 but here that is of the form A' = A^2 + B^2 and B' =0 so the norm is (A^2+B^2)^2 which what is to be expected. What my first post states is that 4Z(Z+1)+1 = (2Z+1)^2 = A + Bi where A = u^2 + v^2 and B = 2uv which I showed in a later post to be true. The product of two conjugates are also of the form A' = u^2+v^2 abd B' = 2uv since this is a trival case where v = 0.
robert Ihnot
#7
Sep24-07, 11:36 AM
P: 1,059
O.K., you have (a+bi)^2 = a^2-b^2 +2abi. Thus N(a+bi)^2 =(a^2-b^2)^2+(2ab)^2 = (A^2+b^2)^2.

So then you are saying (2Z+1)^2 = A+Bi is such that (2Z*+1)^2(2Z+1)^2 = A^2 +B^2.

I guess I was confused about how you were writing that up.
ramsey2879
#8
Sep24-07, 02:00 PM
P: 894
Quote Quote by robert Ihnot View Post
O.K., you have (a+bi)^2 = a^2-b^2 +2abi. Thus N(a+bi)^2 =(a^2-b^2)^2+(2ab)^2 = (A^2+b^2)^2.

So then you are saying (2Z+1)^2 = A+Bi is such that (2Z*+1)^2(2Z+1)^2 = A^2 +B^2.

I guess I was confused about how you were writing that up.
I guess we confused each other. I went back and corrected my last post since A = u^2-v^2 not u^2+v^2; but I don't think I ever said anything about the product of the squares of two conjugates except that by inference they too are of the form A+Bi where A = u^2-v^2 and B =2uv.
ramsey2879
#9
Sep24-07, 06:31 PM
P: 894
It is easily shown that all Gaussian integers that are squares are of the form A+Bi where A=u^2-v^2 and B = -2uv. Therefore all Gaussian integers that are squares have a square norm. But not all Gausian integers that have a square norm are squares since 3 is not a Guassian square but has a square norm and 3*Z^2 has a square norm but likewise is not a square. Is it true that all Gaussian integers that have a square norm are either a Gaussian square or a product of a Gaussian square and an integer which is not a Gaussian square?
Thanks for reply
Edit, I forgot to consider the Gaussian units, "i" is not a Gaussian square so I have to amend my question. Are only the only Gaussian integers that have a square norm either a Gaussian square or the product of i and or an integer and a Gaussian square?
robert Ihnot
#10
Sep25-07, 01:31 AM
P: 1,059
I think the first problem here was The Axiom of Symbolic Stability. It is well known that 4n(n+1) + 1 is a square if n is an integer but if n is a Gaussian integer i.e., 4n(n+1) + 1 = A + Bi, then the norm (A^2 + B^2) is always a square! The proof is quite easy since A = u^2 - v^2 and B = 2uv. I failed to recognize that A+Bi was the same as u+vi.

I also was considering, wrongly, a Gaussian integer to be only those that decomposed over the imaginary. I had not considered 3, for example. With the exception of 2 the only primes that will decompose are those congruent to 1 Mod 4. Thus the Pythagorian triples are built up from 5, 13, 17, etc. For example 5 =(1+2i)(1-2i) =(2+i)(2-i). Or products of primes==1 Mod 4 such as: 65 = 8^2+1^1 = 7^2+4^2. (Which can be done in two distinct ways.) However, 3 can be present only in a squared form, such as: (15)^2=9^2+12^2.

Without going into the question of a Gaussain square, I think you are right. As for 2, (1+i) and (1-i) and not distinct primes since (-i)(1+i) = (1-i), so they differ only by a unit.


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