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Question Number 73295 by naka3546 last updated on 10/Nov/19

$$Howmanysolutionsothat$$$$3n-4,4n-5,5n-13$$$$areprimenumbers?$$

Answered by mind is power last updated on 10/Nov/19

$$(5\mathrm{n}-13).(3\mathrm{n}-4)={15\mathrm{n}}^2-59\mathrm{n}+52$$$$={\mathrm{n}}^2-\mathrm{n}+{14\mathrm{n}}^2-58\mathrm{n}+52=\mathrm{n}(\mathrm{n}-1)+2({7\mathrm{n}}^2-29\mathrm{n}+26)=2\mathrm{k}$$$$\Rightarrow 2\mid (5\mathrm{n}-13)\vee 2\mid (3\mathrm{n}-4)$$$$3\mathrm{n}-4>2\Rightarrow \mathrm{n}>2\mid 5\mathrm{n}-13>2\Rightarrow \mathrm{n}>3$$$$\mathrm{if}\mathrm{n}>3\Rightarrow \mathrm{m}=\min (3\mathrm{n}-4,5\mathrm{n}-13)>2$$$$\mathrm{since}3\mathrm{n}-4\mathrm{and}5\mathrm{n}-13\mathrm{are}\mathrm{prim}>2,\Rightarrow 2\mathrm{can}\mathrm{not}\mathrm{divide}\mathrm{one}\mathrm{of}\mathrm{them}$$$$\Rightarrow \mathrm{n}⩽3$$$$\mathrm{n}=3\Rightarrow 3\mathrm{n}-4=5,4\mathrm{n}-5=7,5\mathrm{n}-13=2\mathrm{all}\mathrm{prime}$$$$\mathrm{over}\mathbb{N}\mathrm{we}\mathrm{have}\mathrm{only}\mathrm{one}\mathrm{solution}\mathrm{n}=3$$$$\mathrm{n}⩽2\Rightarrow 5\mathrm{n}-13<0\mathrm{prim}\mathrm{is}\mathrm{defnd}\mathrm{over}\mathbb{N}$$$$$$

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