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Question Number 73295 by naka3546 last updated on 10/Nov/19
How  many  solution  so  that  3n−4,  4n−5,  5n−13  are  prime  numbers ?
$${How}\:\:{many}\:\:{solution}\:\:{so}\:\:{that} \\ $$$$\mathrm{3}{n}−\mathrm{4},\:\:\mathrm{4}{n}−\mathrm{5},\:\:\mathrm{5}{n}−\mathrm{13} \\ $$$${are}\:\:{prime}\:\:{numbers}\:? \\ $$
Answered by mind is power last updated on 10/Nov/19
(5n−13).(3n−4)=15n^2 −59n+52  =n^2 −n   +14n^2 −58n+52=n(n−1)+2(7n^2 −29n+26)=2k  ⇒2∣(5n−13)∨2∣(3n−4)   3n−4>2⇒n>2∣5n−13>2⇒n>3  if n>3 ⇒m=min (3n−4,5n−13)>2  since  3n−4 and 5n−13 are prim >2  ,⇒2 can not divide one of them  ⇒n≤3  n=3⇒3n−4=5,4n−5=7,5n−13=2 all prime  over N we have only one solution n=3  n≤2⇒5n−13<0 prim is defnd over N
$$\left(\mathrm{5n}−\mathrm{13}\right).\left(\mathrm{3n}−\mathrm{4}\right)=\mathrm{15n}^{\mathrm{2}} −\mathrm{59n}+\mathrm{52} \\ $$$$=\mathrm{n}^{\mathrm{2}} −\mathrm{n}\:\:\:+\mathrm{14n}^{\mathrm{2}} −\mathrm{58n}+\mathrm{52}=\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{2}\left(\mathrm{7n}^{\mathrm{2}} −\mathrm{29n}+\mathrm{26}\right)=\mathrm{2k} \\ $$$$\Rightarrow\mathrm{2}\mid\left(\mathrm{5n}−\mathrm{13}\right)\vee\mathrm{2}\mid\left(\mathrm{3n}−\mathrm{4}\right)\: \\ $$$$\mathrm{3n}−\mathrm{4}>\mathrm{2}\Rightarrow\mathrm{n}>\mathrm{2}\mid\mathrm{5n}−\mathrm{13}>\mathrm{2}\Rightarrow\mathrm{n}>\mathrm{3} \\ $$$$\mathrm{if}\:\mathrm{n}>\mathrm{3}\:\Rightarrow\mathrm{m}=\mathrm{min}\:\left(\mathrm{3n}−\mathrm{4},\mathrm{5n}−\mathrm{13}\right)>\mathrm{2} \\ $$$$\mathrm{since}\:\:\mathrm{3n}−\mathrm{4}\:\mathrm{and}\:\mathrm{5n}−\mathrm{13}\:\mathrm{are}\:\mathrm{prim}\:>\mathrm{2}\:\:,\Rightarrow\mathrm{2}\:\mathrm{can}\:\mathrm{not}\:\mathrm{divide}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them} \\ $$$$\Rightarrow\mathrm{n}\leqslant\mathrm{3} \\ $$$$\mathrm{n}=\mathrm{3}\Rightarrow\mathrm{3n}−\mathrm{4}=\mathrm{5},\mathrm{4n}−\mathrm{5}=\mathrm{7},\mathrm{5n}−\mathrm{13}=\mathrm{2}\:\mathrm{all}\:\mathrm{prime} \\ $$$$\mathrm{over}\:\mathbb{N}\:\mathrm{we}\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{n}=\mathrm{3} \\ $$$$\mathrm{n}\leqslant\mathrm{2}\Rightarrow\mathrm{5n}−\mathrm{13}<\mathrm{0}\:\mathrm{prim}\:\mathrm{is}\:\mathrm{defnd}\:\mathrm{over}\:\mathbb{N} \\ $$$$ \\ $$

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