Co-prime unit fractions

(This solution assumes that if $a$ , $b$ , and $c$ are mutually co-prime, the greatest common factor of all three $a$ , $b$ , and $c$ is $1$ , but the greatest common factor of just two variables $a$ and $b$ may be greater than $1$ .)

The equation $\frac{1}{c} = \frac{1}{a} + \frac{1}{b}$ rearranges to $(a - c)(b - c) = c^2$ .

Let $p = a - c$ and $q = b - c$ , so that $a = p + c$ and $b = q + c$ and $c^2 = pq$ .

Assume $p$ and $q$ have a common factor $f > 1$ , so that $p = fm$ and $q = fn$ . Then $c^2 = pq = f^2mn$ . Since $c^2$ and $f^2$ are both squares, $mn$ must also be a square, so let $mn = k^2$ . Now $c^2 = f^2k^2$ , so $c = fk$ . Then $a = p + c = fm + fk = f(m + k)$ and $b = q + c = fn + fk = f(n + k)$ . But then $a$ , $b$ , and $c$ have a common factor $f > 1$ , which contradicts the given that they are mutually co-prime. Therefore, $p$ and $q$ do not have a common factor and must be relatively prime.

Since $c^2 = pq$ and $p$ and $q$ are relatively prime, $p$ and $q$ must also be squares, so let $p = r^2$ and $q = s^2$ . Then $c^2 = pq = r^2s^2$ , so that $c = rs$ . Then $a = p + c = r^2 + rs$ and $b = q + c = s^2 + rs$ , and $a + b = r^2 + 2rs + s^2 = (r + s)^2$ . Therefore, $a + b$ must be a square number .

(One example includes $a = 3$ , $b = 6$ , and $c = 2$ , because $\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$ , and $a + b = 3 + 6 = 9 = 3^2$ .)

Let:

$a = A S_{ab} S_{ac}$

$b = B S_{ab} S_{bc}$

$c = C S_{ac} S_{bc}$

where $GCD(n,m) = S_{nm}$ and as $GCD(a,b,c) = 1 . S_{nm}$ are all pairwise coprime, and $A,B,C$ are pairwise coprime with each other and with $S_{bc}, S_{ac}, S_{bc}$ respectively.

$\frac1c= \frac1a + \frac1b = \frac{a+b}{ab}$ ,

Hence $a+b = \frac{ab}{c} = \frac{AB S_{ab}^2}{C}$ .

On the RHS $C$ is coprime from the numerator, so hence must be $1$ (LHS is an integer).

$A S_{ab} S_{ac} + B S_{ab} S_{bc} = AB S_{ab}^2$ .

$A S_{ac} + B S_{bc} = AB S_{ab}$ ,

$A$ must then divide $B S_{bc}$ , but it is coprime, so must be $1$ , similarly so must $B$ .

Hence, $a+b = S_{ab} ( S_{ac} + S_{bc} ) = S_{ab}^2$ , so is always a square number.