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Question Number 17272 by RasheedSoomro last updated on 03/Jul/17

$$\mathrm{Determine}\mathrm{two}\mathrm{distinct}\mathrm{primes}\mathrm{p}\mathrm{and}\mathrm{q}$$$$\mathrm{such}\mathrm{that}:$$$$\left(\mathrm{i}\right)\mathrm{p}+\mathrm{q}+1,\mathrm{p}+\mathrm{q}-1,\frac{\mathrm{p}+\mathrm{q}}{2}\in \mathbb{P}\left(\mathrm{All}\mathrm{primes}\right)?$$$$\left(\mathrm{ii}\right)\mathrm{p}+\mathrm{q}+1,\mathrm{p}+\mathrm{q}-1,\frac{\mathrm{p}+\mathrm{q}}{2},\frac{\mathrm{p}-\mathrm{q}}{2}\in \mathbb{P}\left(\mathrm{All}\mathrm{primes}\right)?$$

Commented by prakash jain last updated on 03/Jul/17

$$\mathrm{p}+\mathrm{q}+1\mathrm{is}\mathrm{prime}.$$$$\mathrm{If}p\mathrm{is}6n+1,qcanotbe6m+1$$$$\mathrm{since}p+q+1=6(n+m)+3\notin \mathbb{P}$$$$p=6n+1$$$$q=6m-1$$$$p+q-1=6(n+m)-1\mathrm{possibly}\mathrm{a}\mathrm{prime}.$$$$\frac{p+q}{2}\mathrm{is}\mathrm{a}\mathrm{prime}=3(n+m)\mathrm{is}\mathrm{prime}$$$$\mathrm{not}\mathrm{possible}.$$$$\mathrm{so}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}p\mathrm{amd}q\mathrm{is}3$$$$\Rightarrow q=3$$$$\mathrm{now}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{a}\mathrm{prime}p$$$$4+p\mathrm{is}\mathrm{prime}$$$$2+p\mathrm{is}\mathrm{prime}$$$$\frac{3+p}{2}\mathrm{is}\mathrm{prime}.$$$$\frac{p-3}{2}\mathrm{is}\mathrm{prime}.$$$$p=6n-1,p+4=6n+3\mathrm{not}\mathrm{a}\mathrm{prime}.$$$$p=6n+1,p+2=6n+3\mathrm{not}\mathrm{a}\mathrm{prime}.$$$$\mathrm{There}\mathrm{are}\mathrm{no}2\mathrm{prime}p\mathrm{and}q$$$$\mathrm{which}\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{four}$$$$\mathrm{conditions}.$$$$\mathrm{Please}\mathrm{review}.$$

Commented by RasheedSoomro last updated on 03/Jul/17

$$\left(\mathrm{i}\right)\mathrm{p}+\mathrm{q}+1\Rightarrow 3+7+1=11$$$$\left(\mathrm{ii}\right)\mathrm{p}+\mathrm{q}-1\Rightarrow 3+11-1=13$$$$\left(\mathrm{iii}\right)\frac{\mathrm{p}+\mathrm{q}}{2}\Rightarrow \frac{7+31}{2}=19$$$$\left(\mathrm{iv}\right)\frac{\mathrm{p}-\mathrm{q}}{2}\Rightarrow \frac{19-13}{2}=3$$$$\mathrm{More}\mathrm{than}\mathrm{two}\mathrm{conditions}\mathrm{to}\mathrm{be}$$$$\mathrm{satisfied}\mathrm{simultaneously}\mathrm{is}\mathrm{required}.$$

Commented by RasheedSoomro last updated on 04/Jul/17

$$\mathrm{Lot}\mathrm{of}\mathrm{thanks}\mathrm{for}\mathrm{my}\mathrm{math}-\mathrm{guru}!$$

Commented by RasheedSoomro last updated on 04/Jul/17

$$\mathcal{I}loveyourstrategytoattacktheproblem!$$$$(\mathrm{I}\mathrm{had}\mathrm{forgotten}(6\mathrm{n}+\mathrm{k})-\mathrm{approach},\mathrm{explained}$$$$\mathrm{by}\mathrm{you}\mathrm{in}\mathrm{past}\mathrm{questions})$$$$NowIcanuseitfortwo/threeconditions.$$$$I-e\frac{p+q}{2}\mathrm{\&}\frac{p-q}{2}\mathrm{be}\mathrm{both}\mathrm{prime}\mathrm{etc}$$$$$$

Commented by prakash jain last updated on 05/Jul/17

$$\mathrm{For}\left(\mathrm{i}\right)\mathrm{three}\mathrm{conditons}$$$$\mathrm{p}+\mathrm{q}+1\mathrm{prime}$$$$\mathrm{p}+\mathrm{q}-1\mathrm{prime}$$$$\frac{\mathrm{p}+\mathrm{q}}{2}\mathrm{prime}$$$$\mathrm{case}\mathrm{p}=6\mathrm{m}+1,\mathrm{q}=6\mathrm{n}-1$$$$\mathrm{p}+\mathrm{q}+1\mathrm{not}\mathrm{possible}$$$$\mathrm{case}\mathrm{p}=6\mathrm{m}+1,\mathrm{q}=6\mathrm{n}+1$$$$\frac{\mathrm{p}+\mathrm{q}}{2}\mathrm{not}\mathrm{possible}$$$$\mathrm{so}\mathrm{one}\mathrm{of}\mathrm{p}\mathrm{and}\mathrm{q}\mathrm{is}\mathrm{either}2\mathrm{or}3.$$$$2\mathrm{is}\mathrm{ruled}\mathrm{out}\mathrm{since}\mathrm{there}\mathrm{is}\mathrm{only}$$$$\mathrm{one}\mathrm{even}\mathrm{prime}\mathrm{we}\mathrm{need}\mathrm{sum}\mathrm{of}$$$$\mathrm{of}\mathrm{two}\mathrm{distinct}\mathrm{primes}\mathrm{to}\mathrm{be}\mathrm{even}.$$$$\mathrm{p}=3,q=6n+1$$$$p+q+1=6n+4\notin \mathbb{P}$$$$\mathrm{p}=3,\mathrm{q}=6\mathrm{n}-1$$$$p+q-1=6n+2\notin \mathbb{P}$$$$p=q=2\mathrm{is}\mathrm{only}\mathrm{option}.$$

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