Conference rusure::math

    For integers x and y, the following formula will generate all
    primitive Pythagorean triangles:
    
		(x�-y�, 2xy, x�+y�)

    Squaring any gaussian integer (x + i*y), x, y, integral, leads to the
    formula in .1.

    For much interesting info on Pythagorean triangles, see Beiler's book
    'Recreations in the Theory of Numbers', published by Dover.

    - Jim

    re. .1
    
          Could you list all sets x,y,z each less than 100?
    
    Kostas G.
    

Re .3:

From the formula Peter mentioned, (x^2-y^2, 2xy, x^2+y^2), we have:

x y  a  b  c
- - -- -- --
2 1  3  4  5
3 1  8  6 10
3 2  5 12 13
4 1 15  8 17
4 2 12 16 20
4 3  7 24 25
5 1 24 10 26
5 2 21 20 29
5 3 16 30 34
5 4  9 40 41
6 1 35 12 37
6 2 32 24 40
6 3 27 36 45
6 4 20 48 52
6 5 11 60 61
7 1 48 14 50
7 2 45 28 53
7 3 40 42 58
7 4 33 56 65
7 5 24 70 74
7 6 13 84 85
8 1 63 16 65
8 2 60 32 68
8 3 55 48 73
8 4 48 64 80
8 5 39 80 89
9 1 80 18 82
9 2 77 36 85
9 3 72 54 90
9 4 65 72 97


				-- edp

    How many primitive Pythagorean triples are there less than N?
    (an asymptotic formula should suffice).

re. .4

      Since I am not sure that you gave all the Pythagorean sets
      x,y,z  each less than 100, and you have not specified which 
      are the primitive sets I have duplicated the efford in this
      reply.

      There are just  50  sets,  x,y,z,  each less than 100,
 
      for which   x^2 + y^2 = z^2.

      Only  16  of these sets are primitive.

      We will use the notation:   x - y - z  for the primitive sets and
                                  x,  y,  z  for the none primitive ones.


      5 -  4 -  3         primitive
     10,   8,   6
     13 - 12 -  5         primitive
     15,  12,   9
     17 - 15 -  8         primitive
     20,  16,  12
     25,  20,  15
     25 - 24 -  7         primitive
     26,  24,  10
     29 - 21 - 20         primitive
     30,  24,  18
     34,  30,  16
     35,  28,  21
     37 - 35 - 12         primitive
     39,  36,  15
     40,  32,  24
     41 - 40 -  9         primitive
     45,  36,  27
     50,  40,  30
     50,  48,  14
     51,  45,  24
     52,  48,  20
     53 - 45 - 28         primitive
     55,  44,  33
     58,  42,  40
     60,  48,  36
     61 - 60 - 11         primitive
     65,  52,  39
     65 - 56 - 33
     65,  60,  25
     65 - 63 - 16         primitive
     68,  60,  32
     70,  56,  42
     73 - 55 - 48         primitive
     74,  70,  24
     75,  60,  45
     75,  72,  21
     78,  72,  30
     80,  64,  48
     82,  80,  18
     85,  68,  51
     85,  75,  40
     85 - 77 - 36         primitive
     85 - 84 - 13         primitive
     87,  63,  60
     89 - 80 - 39         primitive
     90,  72,  54
     91,  84,  35
     95,  76,  57
     97 - 72 - 65         primitive


Enjoy,

Kostas G.
<><><><><>

    Some history ...
    
    It has been proved that there are no integer values for x, y, z,
    that satisfy  x^3 + y^3 = z^3  or  x^4 + y^4 = z^4, etc.
    
    Fermat (1601-1665) wrote in the margin of one of his books that
    he had found "a most wonderful proof"  that there were no integers
    that could satisfy the relation:
    
                     n       n       n 
                    x   +   y    =  z
    
    for values of n more than 2, but that  "the margin was too small
    to contain it."
    
    A great number of mathematicians have done a vast amount of work
    in trying to prove Fermat's theorem, and it has been proved for
    a very extensive range of values of  n,  but not for all values.
    
    Does any one remember the proof for   n = 4?
    
    Enjoy,
    
    Kostas G.
    
     
    

    There is a reasonably concise proof in Beiler's "Recreations in
    the Theory of Numbers". In summary, it assumes that such a solution
    exists and then shows how that implies that a smaller solution also
    must exist. Since the solutions are positive integers, no solution
    can exist.

    Pells equation:
    
    
                        2     2
                       x  - qy  = 1        (q <> 0)
    
                       for  x, y > 0
    
    This equation could be called Fermat's equation. However due to
    Euler's mistake in attributing the equation to the English
    mathematician John Pell (1610 - 1685) this equation is called Pell's
    equation.
    
    Solutions to Pell's equations for nonsquare integers  q  satisfying
    1 < q < 30, and  x,y > 0 follow.

        
      q     x     y
    ------------------
      2     3     2
      3     2     1
      5     9     4
      6     5     2
      7     8     3
      8     3     1
     10    19     6
     11    10     3
     12     7     2
     13   649   180
     14    15     4
     15     4     1
     17    33     8
     18    17     4
     19   170    39
     20     9     2
     21    55    12
     22   197    42
     23    24     5
     24     5     1
     26    51    10
     27    26     5
     28   127    24
     29  9801  1820
    
    
Enjoy,
    
Kostas G.
    
    

    Before I forget,
    
    What are the positive solutions of the Pell's equation:
    
                2      2
               x  - 42y  = 1

    
    Enjoy,
    
    Kostas G.
    
    
    
               

    re .-1: x=13, y=2. Care to try for q=61? That's a real toughie.

    re.  .-1
    
              for  x^2 - 42*y^2 = 1
    
              another solution in addition to   x = 13,  y = 2
    
              x = 337,  y = 52.
    
    

    re. .-2
    
                                        2       2
           for Pell's equation (i.e.   x  -  61y  = 1 )

    
           x = 176631049
    and  
           y = 226153980
    
    
    Enjoy,
    
    Kostas G.
    

    re. .11
    
           This has to be the hardest Pellian problem
    
           when   q = 1621, which means find solutions to the equation:
    
    
                 2      2
                x  -  qy  = 1

    
                when  q = 1621.
    
    
    Note:
    
          When q = 1620  then   x = 161 and y = 4
    
          But when  q = 1621    x  has 76 digits and
                                y  has 77 digits
    
    
    Enjoy,
    
    Kostas G.
    
    
    

    Beiler's "Recreations in the Theory of Numbers" gives the 75-digit
    solution to this problem and also the 150+ - digit solution to the
    case Q= 9781. I won't try to transcribe them here. Obviously there
    is no 'worst case' here; solutions just keep getting (randomly) larger,
    with lots of easy cases in between.

    re. .11
    
           another solution to  x^2 - 42y^2 = 1 
    
           in addition to :      x = 13,    y = 2
                                 x = 337,   y = 52
    
           is also               x = 8749,  y = 1350
    
    
    KGG
    

Let (a,b,c) be a primitive pythagorean triple i.e.
    a,b,c are positive integers such that a�+b�=c� and gcd(a,b,c)=1.

Clearly therefore gcd(a,b) = gcd(b,c) = gcd(a,c) = 1		(1)

It can be shown that either a or b must be even and by (1) they cannot both be
even so I will take b to be even. (c > a and c > b but either a > b or b > a is
possible.)

The 'smallest' such triple is (3,4,5).

One may wonder about the set of positive integers t (t > 2) such that for all
primitive pythagorean triples either t|a or t|b or t|c. (t need not divide the
same member of the triple for each triple.)

The answer is {3,4,5}.

Coincidence?

The proofs are elementary. We can in fact be more specific viz. It is always b
that is divisible by 4 and never c that is divisible by 3 (both 3|a and 3|b
occur). 5|a, 5|b and 5|c all occur.

I must admit that I had never noticed these before.

Prove that (c+a)/2 is a perfect square. Hint follows FF.

Prove at the same time that (c-a)/2 is a perfect square.

RE: -1

Most of the proofs follow directly from the fact that:

     a = m**2 - n**2
     
     b = 4mn

     c = m**2 + n**2

The primitive solution which is necessary and sufficient for
all solutions with gcd(a,b,c)=1.

Lew

re .18
    
>Most of the proofs follow directly from the fact that:
    
    Indeed true but proving those relationships becomes bigger than the
    original statement.

re .18
i think you mean b=2mn, not 4mn.

> i think you mean b=2mn, not 4mn.

Yes.  Thanks.

Re: .19

Of course, but this is in most elementary number theory books, and I
didn't want to simply reproduce here.

Lew

from .17
    
Let (a,b,c) be a primitive pythagorean triple i.e.
    a,b,c are positive integers such that a�+b�=c� and gcd(a,b,c)=1.

Clearly therefore gcd(a,b) = gcd(b,c) = gcd(a,c) = 1		(1)

It can be shown that either a or b must be even and by (1) they cannot both be
even so I will take b to be even.
    
Prove that (c+a)/2 is a perfect square.
    
    Given the above it is obvious that a and c are odd and hence that a+c
    and a-c are even and thus that (c+a)/2 and (c-a)/2 are both integers.
    
         c+a   c-a   c�-a�   b�
         --- . --- = ----- = --    and b�/4 is of course a perfect square.
          2     2      4     4
    
    
    
    Now suppose that t|(c+a)/2 and t|(c-a)/2
    
    => 2t|c+a and 2t|c-a
    
    => 2t|2c and 2t|2a
    
    => t|c and t|a
    
    As we know that gcd(a,c) = 1 we can conclude that gcd((c+a)/2,(c-a)/2) = 1
    
    Hence (c+a)/2 and (c-a)/2 are both perfect squares.