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

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

$$ \\ $$$${show}\:{that}\: \\ $$$${m}^{\mathrm{2}} +{m}\:{and}\:\:\mathrm{2}{m}+\mathrm{1}\:{are}\:{prime}\:{betwen} \\ $$$${them}. \\ $$

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

$${for}\:\mathrm{m}\:\epsilon\:\mathbb{Z}^{\ast} −\left\{−\mathrm{1}\right\} \\ $$$$ \\ $$$${pgcd}\left({m}^{\mathrm{2}} +{m}\:;\:\mathrm{2}{m}+\mathrm{1}\right)={pgcd}\left[{m}\left({m}+\mathrm{1}\right)\:;{m}+\left({m}+\mathrm{1}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:={pgcd}\left[{m};\left({m}+\mathrm{1}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={pgcd}\left({m};\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}\:{so}\:{m}^{\mathrm{2}} +{m}\:{and}\:\mathrm{2}{m}+\mathrm{1}\:{are}\:{prime} \\ $$$${betwen}\:{them} \\ $$

Commented by mathocean1 last updated on 20/Dec/20

thank you all

$${thank}\:{you}\:{all} \\ $$

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 ✓

$$\:\:\left({m}^{\mathrm{2}} +{m},\mathrm{2}{m}+\mathrm{1}\right)={d} \\ $$$$\:\:\:\left\{_{{d}\mid\mathrm{2}{m}+\mathrm{1}} ^{{d}\mid{m}^{\mathrm{2}} +{m}} \:\Rightarrow\left\{_{{d}\mid\mathrm{2}{m}^{\mathrm{2}} +{m}} ^{{d}\mid\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{m}} \Rightarrow{d}\mid{m}\right.\right. \\ $$$$\:\:\:\:\left\{_{{d}\mid\mathrm{2}{m}+\mathrm{1}} ^{{d}\mid{m}} \Rightarrow\left\{_{{d}\mid\mathrm{2}{m}+\mathrm{1}} ^{{d}\mid\mathrm{2}{m}} \Rightarrow{d}\mid\mathrm{1}\overset{{d}\in\mathbb{N}} {\Rightarrow}\:{d}=\mathrm{1}\:\checkmark\:\:\:\:\right.\right. \\ $$

Answered by MJS_new last updated on 20/Dec/20

m^2 +m=m×(m+1)=a×b 2m+1=m+(m+1)=a+b c∣(a×b) ⇒ c∣a ∨ c∣b c∣(a+b) ∧ c∣a ⇔ c∣b ⇒ (c∣(a×b) ∧ c∣(a+b) ⇔ c∣a ∧ c∣b) but c∣m ⇔ c∤(m+1) ∀c≠1 ⇒ c=1

$${m}^{\mathrm{2}} +{m}={m}×\left({m}+\mathrm{1}\right)={a}×{b} \\ $$$$\mathrm{2}{m}+\mathrm{1}={m}+\left({m}+\mathrm{1}\right)={a}+{b} \\ $$$${c}\mid\left({a}×{b}\right)\:\Rightarrow\:{c}\mid{a}\:\vee\:{c}\mid{b} \\ $$$${c}\mid\left({a}+{b}\right)\:\wedge\:{c}\mid{a}\:\Leftrightarrow\:{c}\mid{b} \\ $$$$\Rightarrow\:\left({c}\mid\left({a}×{b}\right)\:\wedge\:{c}\mid\left({a}+{b}\right)\:\Leftrightarrow\:{c}\mid{a}\:\wedge\:{c}\mid{b}\right) \\ $$$$\mathrm{but}\:{c}\mid{m}\:\Leftrightarrow\:{c}\nmid\left({m}+\mathrm{1}\right)\:\forall{c}\neq\mathrm{1} \\ $$$$\Rightarrow\:{c}=\mathrm{1} \\ $$

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