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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
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
$${Suppose}\:{that}\:{the}\:{greatest}\:{common}\:{divisor}\:{of} \\ $$$${the}\:{positive}\:{integers}\:{a},{b}\:{and}\:{c}\:{is}\:\mathrm{1}\:{and} \\ $$$$\frac{{ab}}{{a}−{b}}\:=\:{c}\:.\:{Prove}\:{that}\:{a}−{b}\:{is}\:{a} \\ $$$${perfect}\:{square} \\ $$
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 ∴((ab)/(a−b))∉Z [but c∈Z] if one of a,b even and other is odd ((ab)/(a−b))∈Z only a−b=1 [G.C.D of a,b,c is 1] so a−b perfect square.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{a}−\mathrm{b}=\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\mathrm{If}\:\mathrm{a},\mathrm{b}\:\mathrm{both}\:\mathrm{odd}\:\mathrm{then}\:\mathrm{a}−\mathrm{b}=\mathrm{even} \\ $$$$\therefore\frac{\mathrm{ab}}{\mathrm{a}−\mathrm{b}}\notin\mathrm{Z}\:\left[\mathrm{but}\:\mathrm{c}\in\mathrm{Z}\right] \\ $$$$\mathrm{if}\:\mathrm{one}\:\mathrm{of}\:\mathrm{a},\mathrm{b}\:\mathrm{even}\:\mathrm{and}\:\mathrm{other} \\ $$$$\mathrm{is}\:\mathrm{odd}\:\frac{\mathrm{ab}}{\mathrm{a}−\mathrm{b}}\in\mathrm{Z}\:\mathrm{only}\:\mathrm{a}−\mathrm{b}=\mathrm{1} \\ $$$$\left[\mathrm{G}.\mathrm{C}.\mathrm{D}\:\mathrm{of}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{is}\:\mathrm{1}\right] \\ $$$$\mathrm{so}\:\mathrm{a}−\mathrm{b}\:\mathrm{perfect}\:\mathrm{square}. \\ $$
Commented by bemath last updated on 26/Oct/20
yes
$${yes} \\ $$
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)
$$\mathrm{Supoose}\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathrm{N}^{\ast\:} .\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis} \\ $$$$\mathrm{we}\:\mathrm{have}\:\frac{{ab}}{{a}−{b}}\:=\:{c}\Leftrightarrow\frac{\mathrm{a}\left(\mathrm{b}−\mathrm{a}\right)+\mathrm{a}^{\mathrm{2}} }{\mathrm{a}−\mathrm{b}}=\mathrm{c} \\ $$$$\Rightarrow−\mathrm{a}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}−\mathrm{b}}=\mathrm{c}\Rightarrow\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}+\mathrm{c}\right)=\mathrm{a}^{\mathrm{2}} \left(\bigstar\right) \\ $$$$\mathrm{Suppose}\:\mathrm{gcd}\left(\mathrm{a}−\mathrm{b},\mathrm{a}+\mathrm{c}\right)=\mathrm{d}.\mathrm{Then} \\ $$$$\begin{cases}{\mathrm{a}−\mathrm{b}=\mathrm{md}}\\{\mathrm{a}+\mathrm{c}=\mathrm{nd}}\end{cases}\:\left(\ast\right)\mathrm{with}\:\mathrm{gcd}\left(\mathrm{m},\mathrm{n}\right)=\mathrm{1}\left(\mathrm{d},\mathrm{m},\mathrm{n}\in\mathrm{N}^{\ast} \right) \\ $$$$\left(\bigstar\right)\Rightarrow\mathrm{a}^{\mathrm{2}} =\mathrm{mnd}^{\mathrm{2}} \Rightarrow\mathrm{mn}=\left(\frac{\mathrm{a}}{\mathrm{d}}\right)^{\mathrm{2}} =\mathrm{p}^{\mathrm{2}} \left(\mathrm{p}\in\mathrm{N}^{\ast} \right) \\ $$$$\mathrm{and}\:\mathrm{a}^{\mathrm{2}} =\left(\mathrm{pd}\right)^{\mathrm{2}} \Rightarrow\mathrm{a}=\mathrm{pd},\mathrm{mn}=\mathrm{p}^{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$\mathrm{Since}\:\mathrm{m},\mathrm{n}\:\mathrm{are}\:\mathrm{coprime}\:,\mathrm{from}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{infer} \\ $$$$\mathrm{there}\:\mathrm{esixts}\:\mathrm{u},\mathrm{v}\in\mathrm{N}^{\ast} \mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{m}=\mathrm{u}^{\mathrm{2}} ,\mathrm{n}=\mathrm{v}^{\mathrm{2}} .\mathrm{Hence},\mathrm{from} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\ast\right)\mathrm{we}\:\mathrm{infer}\:\mathrm{a}\vdots\mathrm{d},\:\mathrm{b}\vdots\mathrm{d},\mathrm{c}\vdots\mathrm{d} \\ $$$$,\mathrm{but}\:\mathrm{by}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{gcd}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{d}=\mathrm{1},\mathrm{so}\:\mathrm{a}−\mathrm{b}=\mathrm{md}=\mathrm{u}^{\mathrm{2}} .\mathrm{Thus},\mathrm{a}−\mathrm{b}\:\mathrm{is} \\ $$$$\mathrm{perfect}\:\mathrm{square}\:\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

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