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If-gcd-p-q-1-prove-that-gcd-p-p-q-q-p-q-pq-1-Related-to-Q-69939-

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Question Number 70370 by Rasheed.Sindhi last updated on 04/Oct/19
If gcd(p , q)=1,prove that gcd(p(p+q) , q(p+q) , pq)=1 Related to Q#69939
$${If}\:\:{gcd}\left({p}\:,\:{q}\right)=\mathrm{1},{prove}\:{that} \\ $$$$\:\:\:\:\:{gcd}\left({p}\left({p}+{q}\right)\:,\:{q}\left({p}+{q}\right)\:,\:{pq}\right)=\mathrm{1} \\ $$$$\mathrm{R}\boldsymbol{\mathrm{elated}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{Q}}#\mathrm{69939} \\ $$
Commented by mind is power last updated on 03/Oct/19
suppose That a= gcd(p(p+q),q(p+q),pq)≠1 ⇒ ∃ prime number x that x∣a ⇒x∣pq,x∣q(p+q),x∣p(p+q) x∣pq+q^2 ,x∣pq+p^2 ,x∣pq ⇒x∣q^2 ⇒x∣q x is prime x∣p^2 ⇒x∣p (x∣p ∧ x∣q)⇔x∣gcd(p,q)=1⇒x∣1 absurd x is prime ⇒ther exist not prime factor of a∈N⇔(a=1)
$${suppose}\:{That}\:{a}=\:{gcd}\left({p}\left({p}+{q}\right),{q}\left({p}+{q}\right),{pq}\right)\neq\mathrm{1} \\ $$$$\Rightarrow\:\exists\:{prime}\:{number}\:{x}\:{that}\:{x}\mid{a} \\ $$$$\Rightarrow{x}\mid{pq},{x}\mid{q}\left({p}+{q}\right),{x}\mid{p}\left({p}+{q}\right) \\ $$$${x}\mid{pq}+{q}^{\mathrm{2}} ,{x}\mid{pq}+{p}^{\mathrm{2}} ,{x}\mid{pq} \\ $$$$\Rightarrow{x}\mid{q}^{\mathrm{2}} \Rightarrow{x}\mid{q}\:\:{x}\:{is}\:{prime} \\ $$$${x}\mid{p}^{\mathrm{2}} \Rightarrow{x}\mid{p} \\ $$$$\left({x}\mid{p}\:\wedge\:{x}\mid{q}\right)\Leftrightarrow{x}\mid{gcd}\left({p},{q}\right)=\mathrm{1}\Rightarrow{x}\mid\mathrm{1}\:{absurd}\:\:{x}\:{is}\:{prime} \\ $$$$\Rightarrow{ther}\:{exist}\:{not}\:{prime}\:{factor}\:{of}\:{a}\in\mathbb{N}\Leftrightarrow\left({a}=\mathrm{1}\right) \\ $$
Commented by MJS last updated on 03/Oct/19
good, but I think the last line should be x∣p ∧ x∣q ∧ gcd (p, q)=1 ⇒ x=1 because obviously gcd (p, q)=r ∧ x∣p ∧ x∣q ⇒ x≤r
$$\mathrm{good},\:\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{last}\:\mathrm{line}\:\mathrm{should}\:\mathrm{be} \\ $$$${x}\mid{p}\:\wedge\:{x}\mid{q}\:\wedge\:\mathrm{gcd}\:\left({p},\:{q}\right)=\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$$\mathrm{because}\:\mathrm{obviously} \\ $$$$\mathrm{gcd}\:\left({p},\:{q}\right)={r}\:\wedge\:{x}\mid{p}\:\wedge\:{x}\mid{q}\:\Rightarrow\:{x}\leqslant{r} \\ $$
Commented by mind is power last updated on 03/Oct/19
x∣a∧x∣b⇔x∣gcd(a,b) ifx∣gcd(a,b)⇒x∣a∧x∣b ifx∣a∧x∣b⇒x∣gcd(a,b)
$${x}\mid{a}\wedge{x}\mid{b}\Leftrightarrow{x}\mid{gcd}\left({a},{b}\right) \\ $$$${ifx}\mid{gcd}\left({a},{b}\right)\Rightarrow{x}\mid{a}\wedge{x}\mid{b} \\ $$$${ifx}\mid{a}\wedge{x}\mid{b}\Rightarrow{x}\mid{gcd}\left({a},{b}\right) \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 04/Oct/19
Sir, Thanks & Appreciation→you!
$$\mathbb{S}\mathrm{ir},\:\mathbb{T}\mathrm{hanks}\:\&\:\mathbb{A}\mathrm{ppreciation}\rightarrow\mathrm{you}! \\ $$
Answered by $@ty@m123 last updated on 04/Oct/19
If possible let us assume that gcd{p(p+q),q(p+q),pq}=a≠1 Then ∃ m,n,r∈N with gcd(m,n,r)=1 such that p(p+q)=a.m ___(1) q(p+q)=a.n ___(2) pq=a.r ___(3) From (1) &(3) p^2 +ar=am p^2 =a(m−r) ___(4) From (2) &(3) ar+q^2 =an q^2 =a(n−r) ___(5) (4) &(5)⇒p^2 ∣a ∧q^2 ∣a ⇒p∣a ∧q∣a ⇒gcd(p,q)=a≠1 which is a contradiction. ∴ our assumption is wrong. ∴ gcd{p(p+q),q(p+q),pq}=1
$${If}\:{possible}\:{let}\:{us}\:{assume}\:{that} \\ $$$$\:{gcd}\left\{{p}\left({p}+{q}\right),{q}\left({p}+{q}\right),{pq}\right\}={a}\neq\mathrm{1} \\ $$$${Then}\:\exists\:{m},{n},{r}\in\mathbb{N}\:{with}\:{gcd}\left({m},{n},{r}\right)=\mathrm{1} \\ $$$${such}\:{that} \\ $$$${p}\left({p}+{q}\right)={a}.{m}\:\_\_\_\left(\mathrm{1}\right) \\ $$$${q}\left({p}+{q}\right)={a}.{n}\:\:\_\_\_\left(\mathrm{2}\right) \\ $$$${pq}={a}.{r}\:\:\_\_\_\left(\mathrm{3}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\left(\mathrm{3}\right) \\ $$$${p}^{\mathrm{2}} +{ar}={am} \\ $$$${p}^{\mathrm{2}} ={a}\left({m}−{r}\right)\:\_\_\_\left(\mathrm{4}\right) \\ $$$${From}\:\left(\mathrm{2}\right)\:\&\left(\mathrm{3}\right) \\ $$$${ar}+{q}^{\mathrm{2}} ={an} \\ $$$${q}^{\mathrm{2}} ={a}\left({n}−{r}\right)\:\_\_\_\left(\mathrm{5}\right) \\ $$$$\left(\mathrm{4}\right)\:\&\left(\mathrm{5}\right)\Rightarrow{p}^{\mathrm{2}} \mid{a}\:\wedge{q}^{\mathrm{2}} \mid{a} \\ $$$$\Rightarrow{p}\mid{a}\:\wedge{q}\mid{a} \\ $$$$\Rightarrow{gcd}\left({p},{q}\right)={a}\neq\mathrm{1} \\ $$$${which}\:{is}\:{a}\:{contradiction}. \\ $$$$\therefore\:{our}\:{assumption}\:{is}\:{wrong}. \\ $$$$\therefore\:{gcd}\left\{{p}\left({p}+{q}\right),{q}\left({p}+{q}\right),{pq}\right\}=\mathrm{1} \\ $$
Commented by Rasheed.Sindhi last updated on 04/Oct/19
Thank you sir.I appriciate your approach!
$${Thank}\:{you}\:{sir}.{I}\:{appriciate}\:{your}\:{approach}! \\ $$

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