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co-prime unit fractions (Posted on 2020-03-06) Difficulty: 3 of 5
There are 3 positive integers a, b, c such that 1/c=1/a+1/b. If the greatest common divisor of a, b, c is 1, then what type of number must a+b be(e.g square number, cube number, triangular number...)?

No Solution Yet Submitted by Danish Ahmed Khan    
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Possible solution | Comment 3 of 7 |
If 1/c=1/a+1/b, then c(a+b)=ab, and c divides a, or c divides b, or the factors of c are divided among a and b.
Say the factors of c are xy. Then there are 2 cases:

Case 1
Say a=k*c, then
xy(kxy+b)=kxy*b, when kxy = b(k - 1)
i.e. a= b(k-1)
so b= k*c/(k-1)
but if so then (k-1) divides c.
wlog, we can replace y accordingly in the above:
a= k(x(k-1))
b= kx
c=x(k-1)
Then a+b= k^2x
Since x is a factor of a,b, and c, the required condition can only be met if x=1, making k^2x a square.
Case 2
xy(kx+ly)=kx*ly,
so a= x*y(k*x+l*y)/(l*y) = (kx^2)/l+xy
a-c = kx^2/l, and l divides k or x
and b-c = ly^2/k, and k divides l or y.

k and l can be 1, but then xy(x+y)=xy, with x=1-y, and c is not positive, a contradiction.
Say k is smaller than l. If so, then l divides x, and so b = ly also divides c. This would make 1/b larger than 1/c, a contradiction.
k and l can be the same, then a-c = x^2, b-c=y^2, c=xy,  so a+b=(x-y)^2
wlog let y=k-x
Then x(k-x)(x-y)^2=k^2xy, but this is true only for x = {0,k}
The  first case is a contradiction, since xy would be 0.
In the second case, a,b, and c all have k as a factor, failing the required condition.

So a+b can be square, if a,b, and c all have a common factor, but can only be square with a GCD of 1 if c=(k-1), b=k, and a=k(k-1)), a solution equivalent to xdog's.

Edited on March 7, 2020, 11:30 pm
  Posted by broll on 2020-03-07 22:46:27

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