<s>[INST] The sum of numbers being unexpected squares We have the following result:

If $a$, $b$ and $c$ are pairwise coprime integers, such that
$$\frac 1a+\frac 1b=\frac 1c$$
then $a+b$, $a-c$ and $b-c$ are perfect squares.


What I did.
I tried to prove first that $a+b$ is a perfect square.
Multiplying by $abc$ we get
$$c(a+b)=ab.$$
Since $a$ and $c$ are coprime, we get by Gauss' lemma that
$$a\mid a+b.$$
Symmetrically,
$$b\mid a+b.$$
Since $a$ and $b$ are coprime,
$$ab\mid a+b.$$
But we obviously have $a+b\mid ab$, so
$$a+b=ab$$
and $c=1$.
I think this would imply $a=b=2$ which is absurd since $a$ and $b$ are coprime...

The question. Is this going somewhere? What else could I do?
 [/INST] The equation $\frac1a+ \frac1b = \frac1c$ is equivalent to $ab=c(a+b) \implies (a−c)(b−c)=c^2$.
Since $a,b,c$ are positive integers, ${1\over{a}}<{1\over{c}}$ and ${1\over{b}}<{1\over{c}}$. Hence $a>c$ and $b>c$. For each pair of integers $x,y $ satisfying $xy=c^2$, we get $a−c=x$ , $b−c=y$.
$\gcd(a,b) = 1$ thus if a prime $p$ divides both $x,y$ we have $p$ dividing both $a,b$; thus $\gcd(x,y) = 1$. And since $xy=c^2$ we have individually $x = z_1^2$ and $y= z_2^2$.
$a−c=x$ , $b−c=y$. Thus  $a+b = 2c +x+y = (z_1+z_2)^2$. Hence $a+b$ is a prefect square.
In addition we have also proved that $a-c$ and $b-c$ are perfect squares.
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