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

$I f g c d (p, q) = 1, p r o v e t h a t$ $g c d (p (p + q), q (p + q), p q) = 1$ $You can't use 'macro parameter character #' in math mode$
Commented by mind is power last updated on 03/Oct/19

$s u p p o s e T h a t a = g c d (p (p + q), q (p + q), p q) \neq 1$ $\Rightarrow \exists p r i m e n u m b e r x t h a t x ∣ a$ $\Rightarrow x ∣ p q, x ∣ q (p + q), x ∣ p (p + q)$ $x ∣ p q + q^{2}, x ∣ p q + p^{2}, x ∣ p q$ $\Rightarrow x ∣ q^{2} \Rightarrow x ∣ q x i s p r i m e$ $x ∣ p^{2} \Rightarrow x ∣ p$ $(x ∣ p \land x ∣ q) \Leftrightarrow x ∣ g c d (p, q) = 1 \Rightarrow x ∣ 1 a b s u r d x i s p r i m e$ $\Rightarrow t h e r e x i s t n o t p r i m e f a c t o r o f a \in N \Leftrightarrow (a = 1)$
Commented by MJS last updated on 03/Oct/19

$good, but I think the last line should be$ $x ∣ p \land x ∣ q \land \gcd (p, q) = 1 \Rightarrow x = 1$ $because obviously$ $\gcd (p, q) = r \land x ∣ p \land x ∣ q \Rightarrow x ⩽ r$
Commented by mind is power last updated on 03/Oct/19

$x ∣ a \land x ∣ b \Leftrightarrow x ∣ g c d (a, b)$ $i f x ∣ g c d (a, b) \Rightarrow x ∣ a \land x ∣ b$ $i f x ∣ a \land x ∣ b \Rightarrow x ∣ g c d (a, b)$
Commented by Rasheed.Sindhi last updated on 04/Oct/19

$S ir, T hanks & A ppreciation \to you!$
	Answered by $@ty@m123 last updated on 04/Oct/19
	$If possible let us assume that gcd{p(p+q),q(p+q),pq}=a≠1 Then ∃ m,n,r∈N with gcd(m,n,r)=1 such that p(p+q)=a.m ___(1) q(p+q)=a.n ___(2) pq=a.r ___(3) From (1) &(3) p^2 +ar=am p^2 =a(m−r) ___(4) From (2) &(3) ar+q^2 =an q^2 =a(n−r) ___(5) (4) &(5)⇒p^2 ∣a ∧q^2 ∣a ⇒p∣a ∧q∣a ⇒gcd(p,q)=a≠1 which is a contradiction. ∴ our assumption is wrong. ∴ gcd{p(p+q),q(p+q),pq}=1$
	$I f p o s s i b l e l e t u s a s s u m e t h a t$ $g c d {p (p + q), q (p + q), p q} = a \neq 1$ $T h e n \exists m, n, r \in N w i t h g c d (m, n, r) = 1$ $s u c h t h a t$ $p (p + q) = a . m___(1)$ $q (p + q) = a . n___(2)$ $p q = a . r___(3)$ $F r o m (1) & (3)$ $p^{2} + a r = a m$ $p^{2} = a (m - r)___(4)$ $F r o m (2) & (3)$ $a r + q^{2} = a n$ $q^{2} = a (n - r)___(5)$ $(4) & (5) \Rightarrow p^{2} ∣ a \land q^{2} ∣ a$ $\Rightarrow p ∣ a \land q ∣ a$ $\Rightarrow g c d (p, q) = a \neq 1$ $w h i c h i s a c o n t r a d i c t i o n .$ $∴ o u r a s s u m p t i o n i s w r o n g .$ $∴ g c d {p (p + q), q (p + q), p q} = 1$
	Commented by Rasheed.Sindhi last updated on 04/Oct/19

	$T h a n k y o u s i r . I a p p r i c i a t e y o u r a p p r o a c h!$
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