Conference rusure::math

	To get the ball rolling the fact that these triangles are right angled
        means the hypotenuse is a diameter (2p) hence letting the other two
        sides be x and y we have:

         2    2     2
	x  + y  = 4p

        We want the integral solutions for x and y knowing that p is not
        divisible.

        Marc.

     re .1, .2
     
     Right, .0 said the triangles were circumscribed around the
     circle; .1 took the circle as being circumscribed around the
     triangle.
     
     Dan

I think the premise is wrong: there are only two such triangles for each 
prime p.

For any right triangle, it's not too hard to prove that if c is the 
hypotenuse, the radius p of the incircle is (a+b-c)/2, and that for a 
pythagorean triangle, p is an integer. (Draw the radii to each of the three
sides, then the angle bisectors for the acute angles, then compare sides of
the generated triangles. If any side is odd, two of them are, so a+b-c is
divisible by two.) 

It has also been proven elsewhere that all pythagorean triangles are of the 
form {m^2+n^2, m^2-n^2, 2mn} where m and n are integers, m>n>0.
So 
	    m^2-n^2 + 2mn -(m^2+n^2)
	p = ------------------------
			2

	  = n(m-n)
and if p is prime, then there are just two possibilities: n=1 or m-n=1,
so n=(1 or p), and in either case, m=p+1. 

If you plug each of these into the triangle generator above, you find the
average of the six numbers generated to be (p+1)^2. 

Lynn Yarbrough 

> It has also been proven elsewhere that all pythagorean triangles are of the 
> form {m^2+n^2, m^2-n^2, 2mn} where m and n are integers, m>n>0.

	Not so.   Consider {5p, 4p, 3p} (which just happens to be the triad
that you're missing).

Andrew

     >> It has also been proven elsewhere that all pythagorean triangles are of the
     >> form {m^2+n^2, m^2-n^2, 2mn} where m and n are integers, m>n>0.
     
     I think that's how to generate all of the primitive triples
     (i.e., a, b, c, are relatively prime in pairs), with the
     extra condition that one of m, n be even and the other odd.
     
     Dan

Right. I overlooked the prime multiples of the case p=1.

>     I think that's how to generate all of the primitive triples
>     (i.e., a, b, c, are relatively prime in pairs), with the
>     extra condition that one of m, n be even and the other odd.

m and n must be relatively prime.

The reason p must be one of the odd primes is that the case {6,8,10} occurs
for m=3,n=1, so we can't count 2*(m=2,n=1).

Lynn 

     re .7
     
>>   m and n must be relatively prime.
     
     Yes.  On top of the other conditions they must be that, too.
     
     Dan

    Re .5:
    
>> It has also been proven elsewhere that all pythagorean triangles are of the 
>> form {m^2+n^2, m^2-n^2, 2mn} where m and n are integers, m>n>0.
>
>	Not so.   Consider {5p, 4p, 3p} (which just happens to be the triad
>that you're missing).

    Consider  m=2p, n=p 
    
    then   { 5p, 3p, 4p }

>>> It has also been proven elsewhere that all pythagorean triangles are of the 
>>> form {m^2+n^2, m^2-n^2, 2mn} where m and n are integers, m>n>0.
>>
>>	Not so.   Consider {5p, 4p, 3p} (which just happens to be the triad
>>that you're missing).
>
>    Consider  m=2p, n=p 
>    
>    then   { 5p, 3p, 4p }
>
	Not so.   Your approach yields {5p�, 3p�, 4p�}, a completely different
sequence from the one needed.

Regards,
Andrew.

        I think the proper statement of the theorem is that:
        
        All primitive (i.e., a,b,c relatively prime) Pythagorean
        triples are of the form {m^2+n^2, m^2-n^2, 2mn} where m
        and n are relatively prime integers, m>n>0, with one of
        {m,n} even and the other odd.
        
        Conversely given such m and n, then {m^2+n^2, m^2-n^2, 2mn}
        is a primitive Pythagorean triple.
        
        Dan

        Why only primitive triples?
        
        Consider {5p,4p,3p} for p=3.  The number 15 cannot be
        represented as the sum of two squares of integers.  (No
        integer congruent to 3 modulo 4 can be.)
        
        Dan
        

	I wasn't happy with my reply to this one, and I think that the
dialogue becomes a little scrappy, following Lynn's solution.   So here's a
tidied version.

	Suppose that the sides of the triangle have length a,b,c, where c
is the hypoteneuse.   Some simple comparison of triangles shows that:

	c = a+b-2p

	Now, a canonical form for Pythagorean triangles is:

	{k(m�-n�), 2kmn, k(m�+n�)}

	where k, m and n are positive integers, m>n, m & n are coprime, and
m+n is odd.	Any Pythagorean triangle can be uniquely expressed in this
way.

	So here we have:

	p = �k(m�-n� + 2mn -(m�+n�))
	  = kn(m-n)

	Since p is prime, the bag {k,n,m-n} = the bag {p,1,1}.   We have three
possibilities:
	(1)	k = p, m = 2, n = 1:   {a,b,c} = {3p,4p,5p}
	(2)	k = 1, m = p+1, n = 1: {a,b,c} = {p(p+2),2(p+1),p�+2p+2}
	(3)	k = 1, m = p+1, n = p: {a,b,c} = {2p+1,2p(p+1),2p�+2p+1}

	Only (2) & (3) are primitive.   The average of the six sides of these
two primitive triangles is (p+1)�.

	Note: if p is even (ie: 2) then (1) & (2) seem to give the same
triangle {6,8,10}.   (1) is the canonical form, since in (2), m+n is even.
So to get three distinct triangles, p must be odd.

Andrew.

I went search for primitive Pythagorean triples that had the same
hypotenuse.  The smallest were:

 65� =  16� +  63� =  33� +  56�
 85� =  13� +  84� =  36� +  77�
145� =  17� + 144� =  24� + 143�
185� =  57� + 176� = 104� + 153�
205� =  84� + 187� = 133� + 156�
221� =  21� + 220� = 140� + 171�
265� =  23� + 264� =  96� + 247�
305� = 136� + 273� = 207� + 224�

Some hypotenuses occured more that twice.  The eight smallest of these were:

1105� =  47� + 1104� =  264� + 1073� =  576� +  943� =  744� +  817�
1885� = 427� + 1836� =  516� + 1813� =  924� + 1643� = 1003� + 1596�
2405� = 196� + 2397� =  483� + 2356� = 1316� + 2013� = 1564� + 1827�
2465� = 784� + 2337� =  897� + 2296� = 1407� + 2024� = 1504� + 1953�
2665� =  73� + 2664� =  816� + 2537� = 1207� + 2376� = 1824� + 1943�
3145� = 336� + 3127� =  553� + 3096� = 1463� + 2784� = 2184� + 2263�
3445� =  83� + 3444� = 1044� + 3283� = 1596� + 3053� = 2387� + 2484�
3485� = 236� + 3477� =  747� + 3404� = 1276� + 3243� = 2133� + 2756�

And some hypotenuses were shared by 8 primitive Pythagorean triples!
The smallest were:

32045� =   716� + 32037� =  2277� + 31964� =  6764� + 31323� =  8283� + 30956�
       = 15916� + 27813� = 17253� + 27004� = 21093� + 24124� = 22244� + 23067�
40885� =  3636� + 40723� =  8844� + 39917� = 11603� + 39204� = 14893� + 38076�
       = 16236� + 37523� = 19667� + 35844� = 22116� + 34387� = 26093� + 31476�
45305� =   903� + 45296� =  8056� + 44583� = 11816� + 43737� = 14553� + 42904�
       = 20217� + 40544� = 22407� + 39376� = 25984� + 37113� = 31527� + 32536�
58565� =   484� + 58563� = 10604� + 57597� = 15933� + 56356� = 18164� + 55677�
       = 26307� + 52324� = 28797� + 50996� = 33027� + 48364� = 40893� + 41924�

Would anyone care to explain why the the number of primitive Pythagorean triples
that share a hypotenuse is always a power of two?

        Are all of those hypotenii products of 4n+1 primes?
        Of distinct 4n+1 primes?  (I checked half a dozen of
        them.)
        
        A 4n+1 prime factors as (a + bi)(a - bi) for some
        integers a,b, or, equivalently, can be expressed as
        a^2 + b^2.  A 4n+3 prime cannot.  Primitive triples have
        the form a^2 - b^2, 2ab, a^2 + b^2 with gcd(a,b) = 1 and
        exactly one of a,b odd.  Could products of 4n+1 primes
        sometimes have a lot of such representations?
        
        Dan

        re .-2,,
        
>1105� =  47� + 1104� =  264� + 1073� =  576� +  943� =  744� +  817�
        
        1105 = 5 * 13 * 17 is a product of three 4n+1 primes.
        
        5 = (2 + i)(2 - i) or (1 + 2i)(1 - 2i)
        13 = (3 + 2i)(3 - 2i) or (2 + 3i)(2 - 3i)
        17 = (4 + i)(4 - i) or (1 + 4i)(1 - 4i)
        
        There are eight ways to write 1105 as (a+bi)(a-bi) = a^2 + b^2
        that you can get by multiplying three of the above
        factors together:
        
        	from 5:	(2 + i) or (1 + 2i)
        	from 13: (3 + 2i) or (2 + 3i)
        	from 17: (4 + i) or (1 + 4i)
        
        For n prime factors, that's 2^n, but they'll duplicate
        (a^2 + b^2 = b^2 + a^2), so you get 2^(n-1) ways.
        
        [There are more total ways (counting duplicates) when you
        start throwing in the conjugates and/or negation, but it is
        still a power of two (reduced by a factor of a smaller power
        of two when you remove duplicates).]
        
        (2 + i)(3 + 2i)(4 + i)	= 9 + 32i
        (1 + 2i)(3 + 2i)(4 + i)	= -12 + 31i
        (2 + i)(2 + 3i)(4 + i)	= -4 + 33i
        (1 + 2i)(2 + 3i)(4 + i)	= -23 + 24i
        (2 + i)(3 + 2i)(1 + 4i)	= -24 + 23i
        (1 + 2i)(3 + 2i)(1 + 4i)= -33 + 4i
        (2 + i)(2 + 3i)(1 + 4i)	= -31 + 12i
        (1 + 2i)(2 + 3i)(1 + 4i)= -32 - 9i
        
        This translates to the four Pythagorean triples generated
        by (a,b) = (33,4), (32,9), (31,12), and (24, 23)
        
        	a^+b^2		a^2-b^2		2ab		(a,b)
        	------		-------		---		-----
        
        	1105		1073		264		(33,4)
        	1105		943		576             (32,9)
        	1105		817		744		(31,12)
        	1105		47		1104		(24,23)
        
        Why there are no other solutions is left as an exercise
        for the writer. :-)
        
        Dan
        

        Another way of looking at generating the primitive
        triples is to multiply together a choice of a+bi, a-bi
        for each 4n+1 prime [instead of one of a+bi, b+ai].  b+ai
        is just i(a-bi), the multiplication by i doesn't really
        matter, and using conjugates seems "more obvious" than
        swaping a and b.
        
        Dan
        
        p.s. Is 5 * 13 * 17 * 29 * 37 = 1185665 the hypotenuse in
        2^(5-1) = 16 primitive Pythagorean triples?

        re .-1,
        
>p.s. Is 5 * 13 * 17 * 29 * 37 = 1185665 the hypotenuse in
>2^(5-1) = 16 primitive Pythagorean triples?
        
        1185665 is the hypotenuse in 122 triples including the
        trivial one (0, 1185665, 1185665).  Sixteen of them are
        primitive.
        
        Dan