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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.

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    $\begingroup$ $\gcd(a, b, c) = 1$ means that there is no common prime divisor of the three numbers. It is possible a priori that e.g. $\gcd(a, b) > 1$. Also $a, b, c$ may have more than $2$ prime divisors. $\endgroup$
    – WhatsUp
    Mar 7 at 17:01
  • $\begingroup$ this might help $a^2=(a+c)(a-b)$ $\endgroup$
    – Lozenges
    Mar 7 at 17:40
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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.

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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 $$ x= t^2, \; \; y = st, \; \; z = s^2 $$

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  • $\begingroup$ This is likely incomprehensible to most ENT students since arithmetic of quadratic forms is rarely introduced at that level. As for the "added" remark, the "standard" way to do this using unique factorization is to use the theorem I linked in my answer - which yields a one line proof. That theorem is abstracted precisely to handle equations like this (as I mention there, Weil considers it to be the essence of Fermat's method of infinite descent) $\endgroup$ Mar 8 at 8:57
  • $\begingroup$ gcd-based proofs were already given in my link, e.g. $\,\color{#0a0}{y^2 = xz}\Rightarrow $ $(x,y)^2\! = (x^2,\color{#0a0}{y^2}) = \color{#0a0}x(\overbrace{x,\color{#0a0}z}^{\textstyle \color{#c00}{ d }}) = x,\,$ by $\,1=(d,y)\Rightarrow \color{#c00}1 = (d,\color{#0a0}{y^2}) = (d,\color{#0a0}{xz}) \color{#c00}{= d},\,$ by $\,d\mid x\mid xz\ \ $ $\endgroup$ Mar 8 at 9:29
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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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$\,(a\!-\!b)(a\!+\!c)=a^2\,$ & $\, (\color{#c00}{a\!-\!b,a\!+\!c})=1\,$ hence $\,a\!-\!b\,$ is a square.

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  • $\begingroup$ The ${\rm \color{#c00}{gcd}} =1$ else: $\,p\mid \color{#c00}{a-b}\mid a^2\Rightarrow p\mid a,\,$ so $\,p\mid \color{#c00}{a-b,a+c}\Rightarrow p\mid b,c\, $ contra $(a,b,c)=1\ \ $ $\endgroup$ Mar 7 at 18:52

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