Patterns in Pythagoras
 
3C. Even More Patterns in Right Triangles
and Pythagorean Triples
 
 David W. Hansen
© 2008
 
5. Complete Primitive Pythagorean Triples
 
     In any primitive Pythagorean triple (a. b. c),   b + c   =  (2mn)  +  (m2 + n2)   
                                                                                        =     m2 + 2mn + n2  
                                                                                        =      (m + n)2,
                                                                            _____
so b + c is a perfect square,     and                 Ö b + c   = m + n,                                                                       (1)
which is a factor of a, since                 a   =   m2  –  n2  =   (m  +  n)(m – n).                                                     (2)
                                                                                                _____                                           _____
     Now, if m – n  =  1 in  (2)  above, then a  = m  +  n   =   Ö b + c , from  (1)  above.  Thus, Ö b + c = a,
which is so nice that we shall say:
                                                                                         ______
             Any triple (a, b, c) in which m = n+1, and thus Ö  b + c   =  a, is called a complete triple.
            ________________________________________________________________________
 
 Note that m – n = 1 implies that m = n + 1.
 
     Let’s have some fun and construct a table of complete primitive Pythagorean triples sorted in ascending order by their values of m and n.  Perhaps we shall find some interesting patterns!   As stated above, we must have  m – n
= 1 (or m = n + 1) for each of these triples to be complete.
    
                                                                    Table 5  (Complete Triples)
 
                                                                    m     n           a       b       c
                                                            ---------------------------------------------
                                                                     2     1           3       4       5
                                                                     3     2           5     12     13
                                                                     4     3           7     24     25
                                                                     5     4           9     40     41
                                                                     6     5         11     60     61
 
     So, can you see any interesting patterns? Well, we see that in each complete triple, a is an odd integer, and c is one more than b. This is always true for complete triples since m = n + 1, so
 
                                    a   =   m2 – n2   =  (n+1)2  –  n2   =   n2 + 2n +1  –  n2  =   2n + 1,                                   (3)
 
which is an odd integer. The values of a thus form a set of increasing consecutive odd integers starting at 3. 
 
                                     b    =     2mn    =     2(n + 1)(n)       =    2n2 + 2n,                                                             (4)
and                               c    =   m2 + n2   =  (n + 1)2 + n2    =    2n2  + 2n + 1.                                                      (5)
                                          
Comparing (4) and (5) above, we see that indeed, c = b + 1. Thus, in a complete triple (a, b, c),
 
                                            m = n+1,                a = 2n+1,           and              c = b+1.                                         (6)
 
     Are there other patterns? Yes, since a  <  b for each triple in Table 5, this suggests that all complete triples are regular. Now, n is always  ³  1, so
                                                                             n    >   1 / Ö2  »  0.707.                              Squaring both sides of this inequality, we have                                             n2    >   1 / 2,
or                                                                       2n2    >   1.                                              Adding 2n to both sides of this inequality gives us                                      2n2 + 2n   >   2n + 1,
 
which is just                                                          b     >    a                                   from (4) and (3) above, respectively.
 
Thus, a  <  b, and a complete triple is regular. Combining this result with (3), (4), (5), and (6), we have  
 
              Theorem 37
                                                                                                                                          _____
               A primitive Pythagorean triple (a, b, c) is said to be complete if and only if   Ö b + c = a.
               For any complete triple,   m = n+1,   a = 2n+1,   b = 2n2 + 2n,   c = 2n2 + 2n + 1, and  c = b+1.
               In addition, a < b, so all complete triples are regular.
               ____________________________________________________________________________
 
Any more patterns? Look at Table 5 again, taking note of the sums and differences of the a’s and b’s.
 
      In the first complete triple, (3,4,5), the sum of a and b is 7, and in the second triple, (5,12,13), the sum of a and b is 17, which is not much help. However, the difference of b and a in this second triple is 7, the same value as the sum of a and b in the first triple!  Is this just a coincidence?  Let’s see.
 
      In the second triple (5,12,13), the sum of a and b is 5+12 = 17, and in the third triple (7,24,25), the difference between b and a is 24 – 7, which is also 17! Now, this looks good! 
 
     Checking the other triples in Table 5, we find that for every one of the complete triples in the table, the sum of
a and b for any given triple equals the difference of b and a in the next succeeding triple. Thus, for example, the sum of a and b in the fourth triple (9,40,41) is 49, which is equal to the difference of b and a  (60 – 11 = 49)  in the fifth triple (11,60,61).
 
      It appears that if (a1, b1, c1) is any given complete triple in a list of complete triples sorted in ascending order by their a-values, and (a2, b2, c2) is the next succeeding triple in this list, then 
 
                                                                      a1 + b1   =   b2 –  a2.                                                                         (7)
 
    A very interesting way to look at this is to add a2 to both sides of (7) above, giving us
 
                                                                      a1 +   b1   + a2   =   b2.                                                                       (8)
 
Then, in our list of complete triples, (a1, b1, c1) and its successor triple (a2, b2, c2) look like this:
 
                                                                 a             b            c
                                                      --------------------------------------------                  
                                                                . . .          . . .          . . .
                                              
                                                                 a1            b1           c1                
                                                                                                                  ¬    a1 + b1 + a2  = b2
                                                                 a2           b2            c2
                                                        
                                                                 . . .         . . .           . . .
 
 Let’s look at Table 5 again.                                  
   
                                                              Table 5 (Complete Triples)
   
                                                            m     n         a         b         c
                                                      ----------------------------------------------
                                                             2     1         3      4         5
                                                                                                             ¬   3 +  4 + 5  =  12
                                                             3     2         5      12      13
                                                                                                              ¬  5 + 12 + 7  =  24
                                                             4     3          7      24       25
                                                                                                              ¬  7 +  24 + 9  = 40
                                                             5     4         9       40       41
                                                                                                              ¬   9 + 40 + 11 = 60
                                                             6     5       11       60       61
 
Here we can clearly see that for any two successive complete triples, a1 + b1 + a2  =  b2. How very nice.
 
     But this is not all! Take a look at the values of the a’s and c’s for any two successive complete triples in Table 5 as shown below.
                                                            Table 5 (Complete Triples)
 
                                                           m     n         a         b         c
                                                   -----------------------------------------------------
                                                            2     1         3         4         5
                                                                                                              ¬  3 +  5 + 5  = 13
                                                            3     2         5       12       13
                                                                                                              ¬  5 + 13 + 7 = 25
                                                            4     3         7       24       25
                                                                                                              ¬  7 + 25 + 9 = 41
                                                            5     4         9       40       41
                                                                                                              ¬  9 + 41 + 11 = 61
                                                            6     5       11       60       61
 
Thus, we see that for any two successive triples (a1, b1, c1) and (a2, b2, c2) in Table 5,
 
                                                                           a1 + c1 + a2 =   c2.                                                                    (9)
 
We can easily prove (9) if (8) is true, since
 
                                                     a1 + c1 + a2   =   a1 + (b1 + 1)    +   a2
                                                                           =   (a1 + b1 + a2)   +    1
                                                                           =            b2            +    1                   substituting from (8)  
                                                                           =            c2.
Thus,                                         a1 + c1 +   a2   =   c2.                                                                                      (10)
 
     But, is (8) true? To prove (8), we shall try to prove that if R = (a1, b1, c1) is any given complete triple in a list of complete triples sorted in ascending order by their a’s, and S = (a2, b2, c2) is the next succeeding triple in this list, then a1 + b1 + a2 = b2.
 
     Let m1, n1 and m2, n2 be the m and n values used to generate the complete triples  R  =  (a1, b1, c1),  and
S = (a2, b2, c2), respectively. Then, since R and S are both complete, we know that    
 
                                                                    m1 = n1 + 1      and         m2 = n2 + 1,
or                                                                  n1 = m1 - 1      and          n2 = m2 - 1.                                         (11)
 
     First, let’s carry out some preliminary steps.
 
From (3) above, we have                                                 a = 2n+1,
 
so                                                                a1 = 2n1 + 1      and        a2 = 2n2 + 1.                                        (12a)
 
Using (11), we have                                    a1 = 2 (m1 – 1) + 1   =   2m1 – 1,
and                                                               a2 = 2(m2 – 1) + 1    =   2m2 – 1.                                               (12b)
 
     Also from (3) above, we know that the a’s form a set of increasing consecutive odd integers,   
 
so                                                                 a2        =         a1         +     2,                                                        (13)
 
and substituting from (12a), we get         2n2 + 1    =    (2n1 + 1)    +    2,
 
or                                                                   2n2     =          2n1       +    2,
 
so                                                     n2   =   n1 + 1           or           n1   =   n2 - 1                                               (14)
 
    Again, from (13), we have                         a2       =        a1         +    2,     
 
and substituting from (12b), we get          2m2 – 1    =    (2m1 - 1)   +   2,
                                                                       2m2     =      2m1 +    2,
 
and                                        m2   =   m1 + 1           or          m1   =   m2 - 1                                                      (15)
       
Using these preliminary results, we shall now show that   a1 + b1 + a2    =    b2.      Starting with the left-hand side of this equation, we have                            a1 + b1 + a2
     =            m12 – n12         +     2m1n1   +        m22 – n22                               (Theorem 5)
     =     (m1 + n1)(m1 – n1)  +    2m1n1   +    (m2 + n2)(m2 – n2)                       (factoring)
     =         (m1 + n1)(1)         +    2m1n1   +       (m2 + n2)(1)         [ By Theorem 28, m = n + 1, so m – n = 1. Thus,   
                                                                                                              m1 – n1 = 1 and m2 – n2 = 1. ]
     =         m1      +       n1      +         2m1n1           +   m2     +    n2
     =     (m2 – 1) + (n2 – 1) + 2(m2 – 1)(n2 – 1) +  m2  +   n2       [ from (15) and (14) ]
     =      m2 – 1   +  n2 – 1  +  2m2n2  –  2m2  –  2n2  + 2 + m2 + n2
     =      2m2n2   =    b2
 
So, a1 + b1 + a2 = b2, and we have done it! This proof, together with (10) above, gives us
 
        Theorem 38
 
        If (a1, b1, c1) is any given complete triple in a list of complete triples sorted in ascending
        order by their a-values, and (a2, b2, c2) is the next succeeding triple in this list, then  
        a1 + b1 + a2 = b2 and  a1 + c1 + a2 = c2.
        ______________________________________________________________________
 
     Theorem 38 makes it very easy to construct a table of complete triples. To illustrate this, let’s suppose we have the complete triple (5,12,13) and wish to write out quickly and easily the next two successive triples in a list of complete triples sorted in ascending order by their values of a. We simply write down in a row the three values of a, b, and c from the starting triple: namely, 5, 12, and 13, and write three blanks underneath them for the values of the next triple. (See table A below.) Then, we place a 7 for the new a-value under the a-value of 5 since the values of a are increasing consecutively by 2. (See table B below.) Next, we write the sum of the two a-values (5 and 7) and the b-value (12), giving us 5 + 7 + 12 = 24 for the new b-value, as shown in Table C. Lastly, we add 1 to this new b-value and place this number as the new c-value as shown in table D, since c = b + 1. Thus, (7, 24, 25) is the next successive complete triple in our list.
 
            5     12     13                            5     12     13                       5       12     13                        5     12     13
                                                                                                                     
             _      _       _          →              7      _       _           →          7       24       _          →          7     24     25
 
                     A                                            B                                            C                                           D
 
 
Continuing in this manner to get the next triple, we get
 
 
              5     12     13                         5     12     13                        5      12     13                        5     12     13
                                                          
              7     24     25          →           7     24      25           →         7      24     25            →        7     24     25
 
               _      _       _                         9      _        _                         9       40      _                        9     40     41
    
                       E                                          F                                            G                                         H
 

and so (9, 40, 41) is the second successive complete triple after (5, 12, 13). The third successive complete triple after (5, 12, 13) is (11, 60, 61), obtained as shown below.

 
               5     12     13                           5     12     13                      5      12     13                       5     12     13
               7     24     25                           7     24     25                      7      24     25                       7     24     25
               9     40     41            →           9     40     41          →         9      40     41          →         9     40     41
                                                                                       
               _      _       _                          11      _       _                     11       60      _                     11     60     61
 
                                                                            More Patterns?
 
     Let’s use the technique of Theorem 38 to construct a larger table of complete Pythagorean triples sorted in ascending order by their a-values to see if there are any patterns we have missed.
 
                                                                    Table 6 (Complete Triples)
                                                       
                                                                     m     n         a       b       c
                                                                 -----------------------------------------
                                                                      2     1         3        4        5
                                                                      3     2         5      12      13
                                                                      4     3         7      24      25
                                                                      5     4         9      40      41
                                                                      6     5       11      60      61
                                                                      7     6       13      84      85
                                                                      8     7       15    112    113
                                                                      9     8       17    144    145
                                                                    10     9       19    180    181
                                                                    11   10       21    220    221
                                                                    12   11       23    264    265
                                                                    13   12       25    312    313
 
     Notice in Table 6 that the value of n tells us the number of the triple. For example, (11,60,61) is the fifth triple
in Table 6, and 5 is its value of n. The twelfth triple in the table, (25,312,313), has12 as its value of n. In fact, we shall
call the value of n associated with the nth triple in our list, the number of the triple.  It’s easy to find n for the nth
triple (a, b, c) if we know the value of a. Since a = 2n +1, we can find n by solving this equation for n, getting
n = (a- 1)/2. So, we have
 
         The number of the nth complete triple (a, b, c) in a list of complete triples sorted in ascending order
         by their a-values is its value of n. It can be obtained by using the equation  n = (a –1) / 2.
        ____________________________________________________________________________________
   
     Note also that for any given triple in Table 6, the m and n values for the next triple are each one greater than those of the given triple, and the m and n values of the preceding triple are each one less than those of the given triple. Thus, for the triple (an, bn, cn) with m and n values of m and n, the next triple in order,  (an+1, bn+1, cn+1),
will have m and n values of m+1 and n+1, respectively, and the preceding triple,  (an-1, bn-1, cn-1), will have m and n values of m - 1 and n - 1, respectively. This gives us
 
                                           (an - 1, bn - 1, cn - 1),  where   an - 1   =   2(n-1) + 1,
                                                                                          bn - 1   =  2(m-1)(n-1),   and
                                                                                          cn - 1   =   (m-1)2 + (n-1)2,
 
                                                 (an, bn, cn),         where       an    =   2n+1,
                                                                                              bn   =    2mn,    and
                                                                                              cn   =   m2 + n2,
 
                                         (an + 1, bn + 1, cn + 1),  where    an + 1  =  2(n+1) + 1,
                                                                                           bn + 1  =  2(m+1)(n+1),   and
                                                                                           cn + 1  =  (m+1)2 + (n+1)2.
 
     Now, in Table 6 there are several very interesting patterns hidden in this maze of seemingly unrelated data. Can you find any of them? Before reading on, why not see if you can find some.  It’s well worth the effort and lots of fun if you do!    (I’ll wait.)     
 
 
                       -    -    -    -    -    -    -    -    -    -    -    -    -    -    -      -     -    -    -    -    -    -    -    -    -    -
 
    
           Did you find any patterns? One I was fortunate enough to discover is this:
 
     Add any two consecutive b-values in Table 6, such as the first two: 4 + 12 = 16, or the second and third ones: 12 + 24 = 36, or the fifth and sixth ones: 60 + 84 = 144. In each case, the sum is a perfect square! Check it out for yourself  -  the sum of any two consecutive values of b from Table 6 is a perfect square. We have
                                                   b1 + b2   =     4 + 12   =    16   =   42   =   (2x2)2
                                                   b2 + b3   =   12 + 24   =    36   =   62   =   (2x3)2
                                                   b3 + b4   =   24 + 40   =     64  =   82   =   (2x4)2
                                                    b4 + b5  =   40 + 60   =   100  =  102  =   (2x5)2,                        and in general,
                                                         bn - 1  +   bn   =    (2n)2        for n = 2, 3, 4,  . . .
 
But that is not all! Let’s add two consecutive a-values and compare their sum to the corresponding sum of the b-values above. We get
                                  a1 + a2  =   3 + 5  =   8     and     2(a1 + a2)   =   2(8)    =   16   =    b1 + b2,           
                                  a2 + a3  =   5 + 7  = 12,    and     3(a2 + a3)   =  3(12)   =   36   =    b2 + b3,
                                  a3 + a4  =   7 + 9  = 16,    and     4(a3 + a4)   =  4(16)   =   64   =    b3 + b4, 
                                  a4 + a5  =  9 + 11  = 20     and    5(a4 + a5)   =  5(20)   =  100  =    b4 + b5,                and in general                                 n(an - 1­ + an)    =   bn - 1 + bn     =    (2n)2     for n = 2, 3, 4, . . .
Here is a proof.
     Let (an, bn, cn) be the nth triple and (an-1, bn-1, cn-1) be the (n-1)st triple in a list of complete triples
sorted in ascending order by their a’s. Then, since in general, a = 2n + 1 and b = 2mn, we have 
an - 1 = 2(n-1) + 1 = 2n – 1,  an = 2n + 1,   bn - 1  = 2(m – 1)(n – 1), and    bn = 2mn.     Thus,
       1.                        n (an - 1 + an ) = n [ (2n - 1) + (2n + 1) ] = n (4n) = 4n2 = (2n)2.                                     (16)  
       2.                            bn - 1 + bn   =    2(m – 1)(n – 1) + 2mn    =    2(n)(n– 1) + 2(n +1)n 
                                                        = 2(n2 - n + n2 + n) =   2(2n2)   = 4n2   = (2n)2.                                       (17)
Since (16) and (17) are identical, we have thus proved
 
            Theorem 39
 
             If (an, bn, cn) is the nth complete triple and (an - 1, bn - 1, cn - 1) is the preceding triple in
             a list of complete triples sorted in ascending order by their a-values, then
 
                             bn - 1 + bn   =   n(an - 1 + an )   =   (2n)2,    for n   = 2, 3, 4, . . .
             ___________________________________________________________________
 
A very nice theorem!
       Let’s list below some other patterns that I have found and which you also might have discovered. 
Each pattern applies to a set of two successive complete triples, either (an, bn, cn)  and its successor
(an +1, bn + 1, cn + 1), or its predecessor (an - 1, bn - 1, cn - 1), and is listed first and then illustrated using
(9, 40, 41) and (11,60,61), the 4th and 5th triples from Table 6 above.
  
     First, we get the sums, the first two of which we have already proved (Theorem 38).
 
The Sums:
 
1.     an + bn + an+1 = bn+1                       a4       b4      c4
                                                                      9      40      41
        a4 + b4 + a5   = b5
                                                                     a5       b5      c5
         9 + 40 + 11   = 60                               11      60      61
 
 
 
2.     an + cn + an+1   =   cn+1                  a4       b4      c4
                                                                     9       40      41
        a4 + c4 + a5   = c5
                                                                     a5       b5      c5
         9 + 41 + 11 = 61                                 11      60      61
 
 
 
3.    bn + cn+1   =   bn+1 + cn                   a4      b4      c4
                                                                      9      40      41
       b4 + c5   =   b5 + c4
                                                                     a5       b5      c5
      40   + 61   =   60 +   41                         11       60      61          
 
4.   cn + cn+1 = bn + bn+1 + 2             41 + 61 = 40 + 60 + 2
 
5.   bn + bn+1   = (an+1 – 1)2                    40 + 60    =    (11 – 1)2
 
                         =   (an + 1)2                        40 + 60    =      (9 + 1)2
 
                        = [(an + an+1)/2]2                40 + 60    =   [ (9 + 11)/2 ]2
             
 
The Differences:
 
1. For any two consecutive complete triples, the difference between their c-values equals the difference between their b-values,  which equals the sum of their a-values.
  
                             cn+1 – cn   =   bn+1 – bn    =   an+1 + an
                                                                          =   2(an+1 – 1)   =   2(an + 1)
 
                                                                 a4      b4      c4                      
                                                                  9      40      41
                                                  
                                                                 a5       b5      c5                                                                        
                                                                 11      60     61            
                                             
                                                    c5  – c4   =  b5 – b4   = a5 + a4
 
                                                   61  – 41    =  60  – 40 = 9  + 11        
 
                                                                                       = 2(a5 – 1)   =   2(a4 + 1)
 
                                                                                       = 2(11 – 1) =   2(9  + 1)   =    20
 
 
2. The Sum – Product (Theorem 39):   For any two consecutive complete triples, the sum of their b-values equals the sum of their a-values multiplied by the number of the higher triple.
 
                                        bn - 1 +   bn   =   n ( an - 1 + an )   =   (2n)2
 
                                                 m    n         a        b        c
                                              --------------------------------------
                                                 2     1         3        4        5
                                                 3     2         5      12      13
                                                 4     3         7      24      25
                                                 5     4         9      40      41
                                                 6     5       11      60      61
 
                         b1 + b2 = 2(a1 + a2)              4 + 12 = 2(3 + 5) = (2x2)2 = 16
 
                         b2 + b3 = 3(a2 + a3)            12 + 24 = 3(5 + 7) = (2x3)2 = 36
 
                         b3 + b4 = 4(a3 + a4)           24 + 40 = 4(7 + 9) = (2x4)2   = 64
 
                         b4 + b5 = 5(a4 + a5)            40 + 60 = 5(9 +11) = (2x5)2 = 100
 
The Cross-Products:
     Consider now the cross product of the a and b values for any two consecutive complete triples by looking at the first two triples below. Their cross-product is   3(12) – 4(5) = 16 = 42, a perfect square! But that is not all. Add their two b values (4 and 12) together. This sum is equal to the cross-product!     For the second and third triples, we have a cross-product of 5(24) – 7(12) = 36 = 62, another perfect square! And the sum of their two b values (12 + 24) also equals their cross-product. The same is true for the third and fourth triples. There’s certainly something going on here!
                     m    n          a       b          c       anbn + 1 – an + 1bn                     bn + bn+1
                 -------------------------------------------------------------------------------------------------------------
                    2     1          3        4          5
 
                                                                     3(12) – 5(4)   =   36 – 20   = 16 = 4+12 = 42
                                                   
                    3     2          5      12        13
      
                                                                    5(24) – 7(12) = 120 – 84 = 36 = 12+ 24 = 62
 
                    4     3          7      24        25
      
                                                                    7(40) – 9(24) = 280 – 216 = 64 = 24 + 40 = 82
                  
                    5     4          9      40         41
     In general, it appears that
                                 an bn + 1  –   an + 1 bn   =   bn +  bn+1 ,   for n = 0, 1, 2, 3, . . .
Let’s try to prove this. Accordingly,
   an bn + 1  – an + 1 bn   = (2n+1)[ 2(m+1)( n+1)] – [ 2(n+1) + 1] (2mn)
                                  =   2 [ (2n+1)(m+1)( n+1) – 2mn( n+1) – mn ]
                                  =   2 [(2n+1)( mn + m + n + 1) – 2mn2 – 2mn – mn]
                                  =   2[2mn2 + 2mn + 2n2 + 2n + mn + m + n + 1 – 2mn2 – 3mn]
                                  =   2 [2n2 + 3n + m + 1] = 2 [2n2 + 3n + (n+1)+1]
                                  =   2 [2n2 + 4n + 2] = 4(n2 +2n + 1) = 4(n+1)2.                                                              (18)
   bn + bn + 1    =   2mn +2( m+1)( n+1)   = 2(mn + mn + m + n+ 1)
                         =   2[2mn + m + (n+1)] = 2[2mn + m + m] = 2[2mn + 2m]
                         = 4 [mn + m] = 4 [(n+1)n + (n+1)] = 4 [n2 +2n + 1] = 4(n+1)2                                             (19)                
Since (18) and (19) are identical, we have proved              
 1.   an bn + 1 –   an + 1 bn   =   bn + bn+1 = 4(n + 1)2.
 We can also show that              an bn + 1 –   an + 1 bn   =   (n+1)(cn – cn+1)     =        (an + 1)2 
                                                                                     =        (an+1 – 1)2         =   [ (an + an+1) / 2 ] 2
 
 
 
                                                                 a4      b4      c4                      
                                                                  9      40      41
                                                  
                                                                 a5       b5      c5                                                                         
                                                                 11      60     61
                     
                                                 a4b5 – a5b4   =   9(60) – 11(40) = 100
                         
                                                      b4 +  b5     =   40 + 60   =   4(4+1)2
 
                                                    ( n+1)( c5 – c4 ) =   5(61 – 41)
 
                                                             (a4 + 1)2      =       (9 + 1)2
                
                                                            (a5 – 1)2      =       (11 – 1)2
 
                                                    [ (a4 + a5) / 2 ] 2   =   [(9 + 11)/2] 2   = 100
 
 
Here are two more cross-product formulas, one for the a and c values; the other for the b and c values.
2.       ancn+1 – an+1cn   =   (an + 1)2 – 2  
 
                                                                                                              ________                                                    
3.       cnbn+1   – bncn+1   =   bn+1 – bn   = cn+1 – cn   =   2(1 + Ö (1 + 2bn )    =   2(an + 1)
 
 
Those proofs which are not shown for some of the patterns above can easily be proved by simply rewriting the formulas in terms of m’s and n’s, using a = 2n+1, b = 2mn, c= m2 + n2, and m = n+1.
 
     Well, we’re done here for now, and as we can clearly see, there are many patterns to be found.   As Ecclesiastes the Preacher has truly said:
      . . .  of (the) making (of) many books there is no end, and much study is a weariness of the flesh.                         
                                                                                                                                      Ecclesiastes 12:12
Perhaps we should paraphrase this as:
        . . . of (the) making (of) many formulas there is no end, and so many patterns are a weariness to the flesh.
 
Be that as it may, we can certainly say that the field of primitive right triangles and Pythagorean triples is rich in patterns.Perhaps with further exploration and study, we may find other fine patterns whose beauty it will be a joy to experience. Until then, I hope that you have enjoyed reading about the many beautiful patterns we have found here as much as I have enjoyed writing about them. Let me close with the following observation by the brilliant number-theorist Godfrey Harold Hardy.
 
      I believe that mathematical reality lies  outside us, that  our function  is to discover  or observe it,
      and  that the theorems which we prove, and which we describe grandiloquently as our "creations,"
      are simply the notes of our observations.
                                                                                     A Mathematician's Apology ( London 1941 )
 
                                        Part A                                                                                Part B
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