Question

Aratati ca daca numerele p,p²+2sunt prime,atunci p³+2este un numar prim
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Answer 1

$\displaystyle\bf\\\boxed{\bf un~numar~este~prim~daca~are~doar~doi~divizori,~p=prim \Leftrightarrow \mathcal{D}_p=\rightbrace \left\{ 1,p\right\}}~.\\\\\boxed{\bf un~numar~prim~p>3,~poate~lua~doua~forme~:\left \{ {{p=6k+1} \atop {p=6k+5}} \right. }~.\\\\---------------------------------\\\\p~si~p^2+2~sunt~simultan~prime~\implies \boxed{\bf p=prim}~.\\p=2\implies p^2+2=compus.\\\boxed{\bf p=3}~\implies p^2+2=11=prim.\\presupunem~ca~p>3 \implies p~poate~lua~doua~forme~:~ 6k+1~sau~6k+5.$$\displaystyle\bf\\\boxed{\bf cazul~1}~:~p=6k+1 \implies p^2+2=(6k+1)^2+2=36k^2+12k+1+2=\\36k^2+12k+3=3(12k^2+4k+1),~acest~numar~este~prim~daca~\\12k^2+4k+3=1,~de~unde~k=-\frac{1}{3}~(nu~convine),~si~k=0~de~unde~p=3,~solutie~pe~care~deja~o~avem~.\\\\\boxed{\bf cazul~2}~:~p=6k+5 \implies p^2+2=(6k+5)^2+2=36k^2+60k+25+2=\\36k^2+60k+27=3(12k^2+20k+9),~acest~numar~este~prim~daca\\12k^2+20k+9=1,~\implies~nu~avem~solutii~convenabile~.\\prin~urmare~\boxed{\bf p=3}.\\evident,~daca~p=3~atunci~p^3+2=29=prim.$