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.
 A: Suppose $p\mid\gcd(a,b)$, where $p$ is prime. Let $p^r$ be the highest power dividing $a$, and $p^s$ be the highest power dividing $b$. Now since $p\not\mid c$ we must have $p^{r+s}\mid (a-b)$, and in fact it is the highest power dividing $a-b$ (since $(a-b)\mid ab$). If $r>s$, then this is impossible since $p^r\mid a$ but $p^r\not\mid b$, and similarly if $s>r$. So $s=r$.
If $p$ divides exactly one of $a$ and $b$ then it doesn't divide $a-b$. Finally, if $p$ divides neither then it can't divide $a-b$ since $(a-b)\mid ab$.
Thus every prime which divides $a-b$ does so an even number of times.
A: Note $\,(a\!-\!b)(a\!+\!c)=a^2\,$ and $\, \overbrace{(\color{#c00}{a\!-\!b,a\!+\!c})=1}^{\text{see below}}\,$ hence $\,a\!-\!b\,$ is a square. $\ \small\rm QED$
A: the best behaved indefinite ternary quadratic form is $y^2 - zx.$ It takes little to prove that, demanding $x>0$ and $\gcd(x,y,z) = 1,$
$$ x = u^2, \; \; y=uv, \; \; z = v^2   $$
But then all solutions to your $bc-ca+ab=0$   come from
$$ a=x+y, \; \; b = y, \; \; c = y+z . $$
You wanted $a-b.$
Here
$$ a-b = x+y-y = x = u^2  $$
Alright, let me work on the reverse direction a bit
Not bad: In $y^2 - zx = 0,$ take $x=a-b, y = a, z=a+c.$  So that, demanding $x>0,$  it is a square.   We have $\gcd(x,y,z) = \gcd(a,b,c)$  and a little effort tells us this is $1$
ADDED. There is already an answer that refers to unique factorization; here let me talk about $zx=y^2$ using just gcd. The extra demands are that $x >0$  and $\gcd(x,y,z) = 1.$  We begin with $g=gcd(z,y),$  so that $z=gs, y=gt,$ and $\gcd(s,t)=1.$
We reach $gsx = g^2 t^2,$  or $sx = g t^2.$  As  $\gcd(s,t)=1,$  we know that $s|g$  so that we may write $g = su.$  This leads to $sx= sut^2$ and $x=ut^2,$   so that $u>0.$
Next we combine to get $y=stu$ and $z = us^2$
We have reached $$ x=u t^2, \; \; y = stu, \; \; z = u s^2  $$
However, as $u$ divides all three and $\gcd(x,y,z) = 1,$  we know $u=1$  Finally
$$ x= t^2, \; \; y = st, \; \; z =  s^2  $$
A: 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 $p\not \mid m-n$.
