Pythagoras theorem.

[Lior]

Hi,

I would wanna know how to be able to find groups of numbers who are valid groups for the known pythagoras theorem (a^2+b^2=c^2)

I mean, how can I represent A, B, and C with a general variable, let's say n.

so:

A = something with n
B = something with n
C = something with n

so A^2+B^2=C^2

therefore, when I know the number A, I can know also the numbers B and C

of course that in this way, B and C would be only one option for making a group with A.

I hope I made myself clear, sorry, that's my poor english level so far.

[Starman]

I found this which looks a little like you asked for:

The Babylonians were so advanced, they had a theorem for discovering Pythagorian triples.

If p and q take on all whole values subject only to the conditions

1) p > q > 0,
2) p and q have no common divisors other than 1,
3) p and q are not both odd,
then the expressions
x = p2 - q2,
y = 2pq
z = p2 + q2,
will produce all reduced Pythagorean number triples, and each triple only once. (Aaboe 30-31)

From:http://nova.bsuvc.bsu.edu/~00dbwolfe/paper.htm

[jim mcnamara]

Here is an extensive discussion of Pythagorean triples, with several ways to enumerate all of them.

http://mathworld.wolfram.com/PythagoreanTriple.html

Generally, you count only those which consist of relatively prime numbers, since there is a group of infinite arrays that contain all of the infinitely many pythaogrean triples.

[Guv]

Starman: I hope you meant the following.

x = p^2 - q^2 or p*p - q*q

y = 2pq

z = p^2 + q^2 or p*p + q*q

[Starman]

Hi Guv,

I haven't seen you around for a bit.

I did mean that, it's something I found on the web. I just pasted it in and it lost the superscript for the 2's.

Thanks for the correction.

I must check what I have posted.
I must check what I have posted.
I must check what I have posted.
I must check what I have posted.
I must......

[SLH]

Just so you know...

X[sup]2[/sup] produces x²

[Starman]

testing...

X[s<i></i>up]2[/s<i></i>up] produces x²</font></p>

X²

[Starman]

oops!

wrong button

Sorry I should have hit preview.



Thanks SLH, it works.

[DavidHooper]

Oh wow, that's really cool. I never knew that before...

x² x² x² x²
x² x² x² x²
x² x² x² x²

[jim mcnamara]

This is cool too:

x₂

[jim mcnamara]

Oooooops.

[ sub ]2[ /sub ] for a subscript

[nishantp]

Originally posted by Lior
Hi,

I would wanna know how to be able to find groups of numbers who are valid groups for the known pythagoras theorem (a^2+b^2=c^2)

I mean, how can I represent A, B, and C with a general variable, let's say n.

so:

A = something with n
B = something with n
C = something with n

so A^2+B^2=C^2

therefore, when I know the number A, I can know also the numbers B and C

of course that in this way, B and C would be only one option for making a group with A.

I hope I made myself clear, sorry, that's my poor english level so far.

btw heres some VB code that'll do that for you. Make a multiline textbox with a scroll bar, and add this to a button:

Dim A As Long
    Dim B As Long
    Dim D As Long
    Dim N As Long
    Dim M As Long
    
    For M = 1 To 1000
        N = M + 1
        A = Sqr(N) - Sqr(M)
        B = 2 * N * M
        c = Sqr(N) + Sqr(M)
        Text1 = Text1 & A & "²" & vbTab & B & "²" & vbTab & "= " & c & "²" & vbCrLf
    Next

I just applied the formula.

[Guv]

There is no single parameter method for creating general pythagorean triples. The best you can do is something like the formulae Starman posted. Repeated below.

Choose p & q as per the following.

p > q
p & q must have no common factor.
Either p or q (not both) must be odd.

Then

x = 2*p*q
y = p² - q²
z = p² + q²

Using the above, x is sometimes smaller than y & sometimes larger.

Some examples.

(2, 1) results in (3, 4, 5)
(3,2) results in (5, 12, 13)
(8, 3) results in (48, 55, 73)

If (u, v) have a common factor or if both are odd, you get a pythagorean triple with a common factor. For example.

(3, 1) results in (6, 8, 10) or twice (3, 4, 5)

[Guv]

The following use a single parameter, but will not generate all Pythagorean triples as will the (p, q) pairs described in a previous post.

2*n, n² - 1, n² + 1, where n is even.

4*n, n² - 4, n² + 4, where n is odd

The above are based on the previously posted (p, q) formulae.

x = 2*p
y = p² - q²
z = p² + q²

[Guv]

The following use a single parameter, but will not generate all Pythagorean triples as will the (p, q) pairs described in a previous post.

2*n, n² - 1, n² + 1, where n is even.

4*n, n² - 4, n² + 4, where n is odd

The above are based on the previously posted (p, q) formulae.

x = 2*p
y = p² - q²
z = p² + q²

The (p, q) formulae can be used to generate other single parameter formulae.

[smwwilson]

take an odd number... say 5
square it to get 25. take the two numbers closest to each other that add up to 25 (12 and 13). now 5, 12, and 13 are your numbers.

say you have an even number ... say 6. divide it by two until you get an odd number. in this case 3. repeat above steps...
3*3 and get 9. get 4 and 5 because they are the closest numbers to each other that add up to 9. since we divided 6 by 2 only once, multiply 4 and 5 by two to get 8 and 10. now, we have 6 (our original #) and 8 and 10. voila!

powers of 2 dont work as the lowest of the 3 numbers.