show that m^2 +m and 2m+1 are prime betwen them.


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Question Number 126391 by mathocean1 last updated on 20/Dec/20

$s h o w t h a t$ $m^{2} + m a n d 2 m + 1 a r e p r i m e b e t w e n$ $t h e m .$
	Answered by akornes last updated on 20/Dec/20
	$for m ε Z^∗ −{−1} pgcd(m^2 +m ; 2m+1)=pgcd[m(m+1) ;m+(m+1)] =pgcd[m;(m+1)] =pgcd(m;1) =1 so m^2 +m and 2m+1 are prime betwen them$
	$f o r m ϵ Z^{*} - {- 1}$ $p g c d (m^{2} + m; 2 m + 1) = p g c d [m (m + 1); m + (m + 1)]$ $= p g c d [m; (m + 1)]$ $= p g c d (m; 1)$ $= 1 s o m^{2} + m a n d 2 m + 1 a r e p r i m e$ $b e t w e n t h e m$
	Commented by mathocean1 last updated on 20/Dec/20

	$t h a n k y o u a l l$
	Answered by mnjuly1970 last updated on 20/Dec/20
	$(m^2 +m,2m+1)=d {_(d∣2m+1) ^(d∣m^2 +m) ⇒{_(d∣2m^2 +m) ^(d∣2m^2 +2m) ⇒d∣m {_(d∣2m+1) ^(d∣m) ⇒{_(d∣2m+1) ^(d∣2m) ⇒d∣1⇒^(d∈N) d=1 ✓$
	$(m^{2} + m, 2 m + 1) = d$ ${_{d ∣ 2 m + 1}^{d ∣ m^{2} + m} \Rightarrow {_{d ∣ 2 m^{2} + m}^{d ∣ 2 m^{2} + 2 m} \Rightarrow d ∣ m$ ${_{d ∣ 2 m + 1}^{d ∣ m} \Rightarrow {_{d ∣ 2 m + 1}^{d ∣ 2 m} \Rightarrow d ∣ 1 \overset{d \in N}{\Rightarrow} d = 1 ✓$
	Answered by MJS_new last updated on 20/Dec/20

	$m^{2} + m = m \times (m + 1) = a \times b$ $2 m + 1 = m + (m + 1) = a + b$ $c ∣ (a \times b) \Rightarrow c ∣ a \lor c ∣ b$ $c ∣ (a + b) \land c ∣ a \Leftrightarrow c ∣ b$ $\Rightarrow (c ∣ (a \times b) \land c ∣ (a + b) \Leftrightarrow c ∣ a \land c ∣ b)$ $but c ∣ m \Leftrightarrow c ∤ (m + 1) \forall c \neq 1$ $\Rightarrow c = 1$
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