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 }

        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
        

        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

        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