Gcd reciprocals

We may consider cases on the GCDs. In case all $a,b,c$ are coprimes

$a+\frac{1}{b}=3$

There is a solution to it $(2,1)$ , however, $a$ would be greater than $b$ .

As the second case, we may assume $a,b,c$ are coprimes, except for exactly one pair. If the pair is $(a,b) \neq 1$

$a+\frac{1}{b}=2+\frac{1}{(a,b)}$

One can show that $A+\frac{1}{a}=B+\frac{1}{b}$ , if and only if $A=B$ and $a=b$ . Therefore, for the equation above, we have $a=2$ . Consequently, $(a,b) \leq 2 < b$ , since $a<b$ . So, $b \neq (a,b)$ . Similar steps can be taken, for the case where $(a,c) \neq 1$ is the only pair that is not coprime, to show no such integers exist.

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For the case $(b,c) \neq 1$ , we have

$a+\frac{1}{b}=2+\frac{1}{(b,c)}$

then $a=2$ and if $c=bk$ , for some positive integer $k$ , then $b=(b,c)$ , which is exactly what we need :)

Possible $(b,c)$ is only $(3,9)$ , therefore $(2,3,9)$ is one solution.

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In case exactly one of the pairs of $a,b,c$ is coprime, then $a=1$ and, hence, $(a,b)=1$ and $(a,c)=1$ . Since at least one of the mentioned GCDs appear in the equation

$a+\frac{1}{b}=\frac{1}{(a,b)}+\frac{1}{(a,c)}+\frac{1}{(b,c)}$

then $\frac{1}{b}$ should become a fraction greater than one, chichis not possible.

As the last case, we consider that all the pairs are not coprime. Then the following maximum can be deduced

$\frac{1}{(a,b)}+\frac{1}{(a,c)}+\frac{1}{(b,c)} \leq \frac{1}{2}+\frac{1}{4}+\frac{1}{6} < 1$ . But we know $a+\frac{1}{b} > 1$ , and we are done.

We immediately notice that an equation like this does not have much algebraic structure we can use to directly solve for the variables (This is because the $\gcd$ lacks useful algebraic identities). Thus, we turn to inequalities. Notice that the left hand side cannot be bigger than $3$ , this is because the minimum value of the greatest common divisor of two numbers is $1$ so we deduce: $a + \frac{1}{b} \leq 3 \implies ab + 1 \leq 3b \implies b(a-3) \leq -1$ . From this, we obtain that $a < 3$ because otherwise, as $b$ is positive, the left-hand side will be non-negative. So the only possible values for $a$ are $2,1$ but it is clear that $a \neq 1$ because if that was the case $1 + \frac{1}{b} = 2 + \frac{1}{(b,c)} \implies \frac{1}{b} = 1 + \frac{1}{(b,c)} \implies \frac{1}{b} > 1$ which is a clear contradiction. In other words, we have deduced that $a = 2$ .

The equation is now $2 + \frac{1}{b} = \frac{1}{(2,b)} + \frac{1}{(2,c)} + \frac{1}{(b,c)}$ . But we know that $b \geq 3$ so we have the double inequality $2 + \frac{1}{3} \geq \frac{1}{(2,b)} + \frac{1}{(2,c)} + \frac{1}{(b,c)} > 2$ which turns out to be a very tight bound for our purposes. We will use this inequality to prove that both $b,c$ have to be odd. If neither is odd, both are even so we would have $2 + \frac{1}{3} \geq 1 + \frac{1}{(b,c)} > 2 \implies 1 + \frac{1}{3} \geq \frac{1}{(b,c)} > 1$ but this is a clear contradiction because $\frac{1}{(b,c)} \leq 1$ . So at least one of them has to be odd. But then if one is odd and the other even, we would have $2 + \frac{1}{3} \geq \frac{3}{2} + \frac{1}{(b,c)} > 2 \implies \frac{2}{3} \geq \frac{1}{(b,c)} > \frac{1}{2}$ but no rational number of the form $\frac{1}{n}$ satisfies this inequality. We can see this because $\frac{1}{n}$ does not satisfy the rightmost inequality for $n \geq 2$ , and it does not satisfy the leftmost inequality for $n = 1$ . In other words, a contradiction. So both $b,c$ are odd.

So we now have to solve the equation $2 + \frac{1}{b} = 2 + \frac{1}{(b,c)} \implies \frac{1}{b} = \frac{1}{(b,c)} \implies b = (b,c)$ . This is, finally, an equation that is very susceptible to typical algebraic techniques. This immediately says that $c$ has to be a multiple of $b$ . So $c = kb$ where $k \geq 3$ because $k$ has to be an odd integer bigger than 1 (if it is equal to 1 then $b = c$ which is a contradiction) and $b \geq 3$ because $b > a$ . Checking, $b = k = 3$ works. But then if either $b,k$ is bigger than 3, we have that $bk \geq 3 \times 5 = 15$ which is a contradiction. So the only solution is $a=2, b=3, c=9$ .

However, it is more interesting to note that if we ignore the upper bound on the variables, this equation has general solution $(2, 2n + 1, (2n +1)(2m + 1)), m,n \in \mathbb{N}$ .