<s>[INST] If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. $\\$
Well I tried expressing $a=p_1^{a_1}.p_2^{a_2} \cdots p_k^{a_k}$ and $b = q_1^{b_1}.q_2^{b_2}\cdots q_k^{b_k}$ and $c=r_1^{c_1}.r_2^{c_2}\cdots r_k^{c_k}$ basically emphasizing on the fact that the primes which divide $a$ are different from those that divide $b$ and $c$, but I couldn't come up with anything fruitful. $\\$
Any help would be appreciated. $\\$
Thanks
EDIT:- $a,b,c$ are positive integers.
 [/INST] Well $\frac {ab}{a-b}$ being an integer seems unlikely and specific.
In particular if $p$ is a prime divisor of $c$ then $p|\frac {ab}{a-b}$ so $p|ab$ so $p|a$ or $b|b$.  But $p$ can't divide both as $\gcd(a,b,c) =1$.
But we have that if $p|c$ then either $p|a$ or $p|b$ but not both but then $p\not \mid a-b$ yet for any $q|a-b$ we must have $q|a$ or $q|b$ so $q$ must divide both $a$ and $b$ so....
Okay.... Let $\gcd(a,b) = d$ and with $a=a'd; b= b'd$ and as $\gcd(a,b,c) = 1$ we have $\gcd(d,c) =1$. (Also we have $\gcd(a',b') = 1$.   so.....
$c = \frac {ab}{a-b} = \frac {a'b'd^2}{d(a'-b')} = \frac {a'b'}{a'-b'}d$.
But two things to note.  1) $d\not \mid c$ so we must have $d|(a'-b')$ and $\gcd(a',b') = 1$ so $a' - b'$ can't have any factors in common with $a'$ or $b'$ and so cant have any factors in common with $a'b'$[1].  So we must have $a'-b'|d$.
So we must have:
$c = a'b'\frac d{a'-b'} = a'b'$ and $a'-b' = d$ and $a-b = d(a'-b') = d^2 = \gcd(a,b)^2$.
Too find a case where this is possible we can let $a',b'$ but any two relatively prime integers.  Say $a' = 7$ and $b'=3$ so $a'-b' = 4$.  Multiply both by $4$ to get $a = 28$ and $b = 12$ and let $c = \frac {ab}{a-b} = a'b'$ or $c = \frac {28\cdot 12}{28-12}=\frac {28\cdot 12}{16} = 7\cdot 3=21$.
SO $\gcd(28,12, 21) = 1$ and $21 = \frac {28\cdot 12}{28-12}$.
Cute.
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[1] A cute, but distractingly tangential lemma:  If $\gcd(m,n) =1$ then $\gcd(mn, m- n)=1$.  Pf: If $p|mn$ then $p|m$ or $p|n$  but not both.  So $p\not \mid m-n$.
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