Suppose that the greatest common divisor of the positive integers a,b and c is 1 and ((ab)/(a−b)) = c . Prove that a−b is a perfect square


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Question Number 119657 by bemath last updated on 26/Oct/20

$S u p p o s e t h a t t h e g r e a t e s t c o m m o n d i v i s o r o f$ $t h e p o s i t i v e i n t e g e r s a, b a n d c i s 1 a n d$ $\frac{a b}{a - b} = c . P r o v e t h a t a - b i s a$ $p e r f e c t s q u a r e$
Commented by som(math1967) last updated on 26/Oct/20

$I think a - b = 1 is a perfect square$ $If a, b both odd then a - b = even$ $∴ \frac{ab}{a - b} \notin Z [but c \in Z]$ $if one of a, b even and other$ $is odd \frac{ab}{a - b} \in Z only a - b = 1$ $[G . C . D of a, b, c is 1]$ $so a - b perfect square .$
Commented by bemath last updated on 26/Oct/20

$y e s$
	Answered by 1549442205PVT last updated on 26/Oct/20
	$Supoose a,b,c∈N^(∗ ) .From the hypothesis we have ((ab)/(a−b)) = c⇔((a(b−a)+a^2 )/(a−b))=c ⇒−a+(a^2 /(a−b))=c⇒(a−b)(a+c)=a^2 (★) Suppose gcd(a−b,a+c)=d.Then { ((a−b=md)),((a+c=nd)) :} (∗)with gcd(m,n)=1(d,m,n∈N^∗ ) (★)⇒a^2 =mnd^2 ⇒mn=((a/d))^2 =p^2 (p∈N^∗ ) and a^2 =(pd)^2 ⇒a=pd,mn=p^2 (1) Since m,n are coprime ,from (1)we infer there esixts u,v∈N^∗ so that m=u^2 ,n=v^2 .Hence,from (1) and (∗)we infer a⋮d, b⋮d,c⋮d ,but by the hypothesis gcd(a,b,c)=1 ⇒d=1,so a−b=md=u^2 .Thus,a−b is perfect square (q.e.d)$
	$Supoose a, b, c \in N^{} . From the hypothesis$ $we have \frac{a b}{a - b} = c \Leftrightarrow \frac{a (b - a) + a^{2}}{a - b} = c$ $\Rightarrow - a + \frac{a^{2}}{a - b} = c \Rightarrow (a - b) (a + c) = a^{2} (★)$ $Suppose \gcd (a - b, a + c) = d . Then$ ${\begin{cases} a - b = md \\ a + c = nd \end{cases} () with \gcd (m, n) = 1 (d, m, n \in N^{})$ $(★) \Rightarrow a^{2} = {mnd}^{2} \Rightarrow mn = {(\frac{a}{d})}^{2} = p^{2} (p \in N^{})$ $and a^{2} = {(pd)}^{2} \Rightarrow a = pd, mn = p^{2} (1)$ $Since m, n are coprime, from (1) we infer$ $there esixts u, v \in N^{} so that$ $m = u^{2}, n = v^{2} . Hence, from$ $(1) and () we infer a ⋮ d, b ⋮ d, c ⋮ d$ $, but by the hypothesis \gcd (a, b, c) = 1$ $\Rightarrow d = 1, so a - b = md = u^{2} . Thus, a - b is$ $perfect square (q . e . d)$
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