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Question Number 162674 by Mathematification last updated on 31/Dec/21

Answered by mindispower last updated on 31/Dec/21

d∣2(3n+1)−3(2n+1)⇒d∣−1 d=1 or use bizou identities −2(3n+1)+3(2n+1)=1 ⇒(2n+1),(3n+1) are prime

$${d}\mid\mathrm{2}\left(\mathrm{3}{n}+\mathrm{1}\right)−\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)\Rightarrow{d}\mid−\mathrm{1} \\ $$$${d}=\mathrm{1} \\ $$$${or}\:{use}\:{bizou}\:{identities} \\ $$$$−\mathrm{2}\left(\mathrm{3}{n}+\mathrm{1}\right)+\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{2}{n}+\mathrm{1}\right),\left(\mathrm{3}{n}+\mathrm{1}\right)\:{are}\:{prime} \\ $$

Answered by mr W last updated on 31/Dec/21

gcd(2n+1,3n+1)=1 prove: if gcd(2n+1,3n+1)=k≠1 then there exist integers p, q such that 2n+1=kp 3n+1=kq ⇒n=((kp−1)/2)=((kq−1)/3) ⇒3kp−3=2kq−2 ⇒(3p−2q)k=1 ⇒3p−2q=(1/k)≠integer since k≠1 but 3p−2q is integer. ⇒contradiction! ⇒k=1

$${gcd}\left(\mathrm{2}{n}+\mathrm{1},\mathrm{3}{n}+\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$$${prove}: \\ $$$${if}\:{gcd}\left(\mathrm{2}{n}+\mathrm{1},\mathrm{3}{n}+\mathrm{1}\right)={k}\neq\mathrm{1} \\ $$$${then}\:{there}\:{exist}\:{integers}\:{p},\:{q}\:{such}\:{that} \\ $$$$\mathrm{2}{n}+\mathrm{1}={kp} \\ $$$$\mathrm{3}{n}+\mathrm{1}={kq} \\ $$$$\Rightarrow{n}=\frac{{kp}−\mathrm{1}}{\mathrm{2}}=\frac{{kq}−\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{kp}−\mathrm{3}=\mathrm{2}{kq}−\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{3}{p}−\mathrm{2}{q}\right){k}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{p}−\mathrm{2}{q}=\frac{\mathrm{1}}{{k}}\neq{integer}\:{since}\:{k}\neq\mathrm{1} \\ $$$${but}\:\mathrm{3}{p}−\mathrm{2}{q}\:{is}\:{integer}.\: \\ $$$$\Rightarrow{contradiction}!\:\Rightarrow{k}=\mathrm{1} \\ $$

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