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Reciprocal Equation #2 (Posted on 2003-07-20) Difficulty: 3 of 5
Find all sets of positive integers A, B, and C which satisfy

1/A = 1/B + 1/C.

See The Solution Submitted by Brian Smith    
Rating: 3.4000 (5 votes)

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solution | Comment 12 of 15 |
solution:
B,C are equal even no's (say 2k each)
A is half of either B or C.

proof:
case 1- B odd & C odd
on taking LCM and adding you'll get sum of two odd numbers in the numerator, which is even and product of two odds in the denom, which is odd.
Thus the result shall be a non integer.

case 2- B odd & C even
on taking the LCM and adding you'll get an even no. above B and odd above C. result: sum is odd. the denominator shall be even. hence non integer result.

case 3- B even and C even but both contain unequal powers of two.
on taking the lower power of 2 common from both numbers in the denominator this becomes a case 2 problem.

case 4 - B and C are equal even numbers(say 2k each).
1/A = 1/2k + 1/2k
= 2/2k
= 1/k
hence A = k.



  Posted by spinoza on 2003-07-23 17:01:56
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