Equation : 1/a = 1/b + 1/c

 
 
1/a \; = \; 1/b \; + \; 1/c

Let   b \; = \; a \; + \; k

Then

1/a \; - \; 1/(a+k) \; = \; 1/c

c \, k \; = \; a \, (a+k)

c \; = \; a^2/k \; + \; a

c   is an integer when   k   is a factor of   a^2

So all triples   (a, b, c)   can be expressed as   (a, \; a+k, \; a^2/k + a)

with   a^2 \; \equiv \; 0 \; \pmod k

 

for example,
 

k = 1   …..   a = n,     b = n + 1,     c = n(n + 1)
k = 2   …..   a = 2n,   b = 2n + 2,   c = 2 n(n + 1)
k = 3   …..   a = 3n,   b = 3n + 3,   c = 3 n (n+1)
k = 4   …..   a = 2n,   b = 2n + 4,   c = n (n+2)
k = 5   …..   a = 5n,   b = 5n + 5,   c = 5 n (n+1)
k = 6   …..   a = 6n,   b = 6n + 6,   c = 6 n (n+1)
k = 7   …..   a = 7n,   b = 7n + 7,   c = 7 n (n+1)
…………………….

with   n \neq 0    and    n \neq -1

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Equation : 1/a = 1/b + 1/c

  1. Eugene says:

    The following formulas also give all solutions :
    $b=tx(x+y), c=ty(x+y), a=txy$, where $t,x,y$ – arbitrary integer numbers.

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