Diophantine Reciprocal Sums

           Diophantine Reciprocal Sums

 

 

          Define reciprocal addition [1/+] thus:

          c      =    a  [1/+]  b    

          if and only if

          1/c   =    1/a  +  1/b  

          if and only if

          c      =    (ab) / (a+b)

 

 

Theorem:

If p and q are such that     pq   =  c^2

Then (c+p) [1/+] (c+q)  =  c

 

          Proof:

(c+p) [1/+] (c+q) 

=  (c+p)(c+q)  /  (c+p+c+q)

=  (c^2+pc+qc+pq)  /  (2c+p+q)

=  (c^2+pc+qc+c^2)  /  (2c+p+q)

=  (2c^2+(p+q)c)  /  (2c+p+q)

=  c(2c+(p+q))  /  (2c+p+q)

          =  c

          Example: c=10;

          c^2 = 100 = 1*100 = 2*50 = 4*25 = 5*20 = 10*10

Therefore:

10  =  11 [1/+] 110 

=  12 [1/+] 60

=  14 [1/+] 35

=  15 [1/+] 30

=  20 [1/+] 20

 

Example: c=9;

          c^2 = 81 = 1*81 = 3*27 = 9*9

Therefore:

9    =   10 [1/+] 90    =   12 [1/+] 36    =   18 [1/+] 18

 

          Example: c=6;

          c^2 = 36 = 1*36 = 2*18 = 3*12 = 4*9 = 6*6

Therefore:

6  =  7 [1/+] 42 

=  8 [1/+] 24

=  9 [1/+] 18

=  10 [1/+] 15

=  12 [1/+] 12

 

          Example: c=30;

          c^2 = 900 = 1*900 = 2*450 = 3*300 = 4*225 = 5*180  = 6*150 = 9*100 = 10*90 = 12*75 = 15*60 = 18*50 = 20*45 = 30*30

Therefore:

30  =  31 [1/+] 930 

=  32 [1/+] 480

=  33 [1/+] 330

=  34 [1/+] 255

=  35 [1/+] 210

=  36 [1/+] 180

=  39 [1/+] 130

=  40 [1/+] 120

=  42 [1/+] 105

=  45 [1/+] 90

=  48 [1/+] 80

=  50 [1/+] 75

=  55 [1/+] 66

=  60 [1/+] 60