Egyptian fraction of a unit fraction
If a and b are distinct positive integers satisfying 3 1 = a 1 + b 1 , find a + b .
Bonus : Generalize it for any unit fraction c 1 , where c is a positive integer.
The answer is 16.
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2 solutions
4+12=16
Note:If c isn't prime,then it will have other possible solutions
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i was doing a summation problem (telescoping sum) which was the sum of 1/n(n+1)
you know this is equal to (1/n) - 1/(n+1) i added 1/(n+1) to both sides and got
n 1 = n + 1 1 + n ( n + 1 ) 1
Clearly, a , b > c If we set, a = c + x , b = c + y for some integers x , y > 0 , the equation becomes, c 1 = c + x 1 + c + y 1 Simplifying, we have, ( c + x ) ( c + y ) = c ( 2 c + x + y ) This becomes, c 2 = x y In general, x + y cannot be unique unless c = p for some prime p , giving, x = p 2 , y = 1 as x = y .
Then we have, a = c 2 + c , b = c + 1 Here, c = 3 , so a + b = 1 2 + 4 = 1 6