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Question Number 80262 by Power last updated on 01/Feb/20

Commented by MJS last updated on 01/Feb/20

are there prime un−integers?!?

$$\mathrm{are}\:\mathrm{there}\:\mathrm{prime}\:\mathrm{un}−\mathrm{integers}?!? \\ $$

Commented by mind is power last updated on 02/Feb/20

p must bee prime ≥3 in this case let P={a∈N ∣ a is prime} a^p +1=(a+1)(Σ_(k=0) ^(n−1) (−1)^(n−k) a^k ) if a=1⇒((a^p +1)/(a+1))=1∉P if a≥2 ⇒ ((a^p +1)/(a+1))=a^(p−1) −a^(p−2) +.........+1≥a^(p−1) +a^(p−3) +........+1 claim p(k) 2^(k−1) ≥k ∀k∈N^∗ for k=1 2^0 ≥1 true Equslity ,∀k∈N assum p(k) true 2^k =2.2^(k−1) ≥2(k)=2k ⇒2^k ≥2k=k+k k≥1⇒2^k ≥k+1 P(k)⇒p(k+1) since P(k) true ⇒∀k∈N^∗ 2^(k−1) ≥k a^(p−1) ≥2^(p−1) ≥p⇒ ((a^p +1)/(a+1)) ≥p +1>p

$${p}\:\:{must}\:{bee}\:{prime}\:\geqslant\mathrm{3} \\ $$$${in}\:{this}\:{case} \\ $$$${let}\:{P}=\left\{{a}\in\mathbb{N}\:\:\:\mid\:{a}\:{is}\:{prime}\right\} \\ $$$${a}^{{p}} +\mathrm{1}=\left({a}+\mathrm{1}\right)\left(\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{n}−{k}} {a}^{{k}} \right) \\ $$$${if}\:{a}=\mathrm{1}\Rightarrow\frac{{a}^{{p}} +\mathrm{1}}{{a}+\mathrm{1}}=\mathrm{1}\notin{P} \\ $$$${if}\:{a}\geqslant\mathrm{2}\:\Rightarrow \\ $$$$\frac{{a}^{{p}} +\mathrm{1}}{{a}+\mathrm{1}}={a}^{{p}−\mathrm{1}} −{a}^{{p}−\mathrm{2}} +.........+\mathrm{1}\geqslant{a}^{{p}−\mathrm{1}} +{a}^{{p}−\mathrm{3}} +........+\mathrm{1} \\ $$$${claim}\:\:{p}\left({k}\right)\:\mathrm{2}^{{k}−\mathrm{1}} \geqslant{k}\:\:\forall{k}\in\mathbb{N}^{\ast} \\ $$$${for}\:{k}=\mathrm{1}\:\mathrm{2}^{\mathrm{0}} \geqslant\mathrm{1}\:{true}\:{Equslity}\:,\forall{k}\in\mathbb{N}\:{assum}\:{p}\left({k}\right)\:{true} \\ $$$$\mathrm{2}^{{k}} =\mathrm{2}.\mathrm{2}^{{k}−\mathrm{1}} \geqslant\mathrm{2}\left({k}\right)=\mathrm{2}{k} \\ $$$$\Rightarrow\mathrm{2}^{{k}} \geqslant\mathrm{2}{k}={k}+{k} \\ $$$${k}\geqslant\mathrm{1}\Rightarrow\mathrm{2}^{{k}} \geqslant{k}+\mathrm{1} \\ $$$${P}\left({k}\right)\Rightarrow{p}\left({k}+\mathrm{1}\right)\:{since}\:{P}\left({k}\right)\:{true}\:\Rightarrow\forall{k}\in\mathbb{N}^{\ast} \:\:\mathrm{2}^{{k}−\mathrm{1}} \geqslant{k} \\ $$$$ \\ $$$${a}^{{p}−\mathrm{1}} \geqslant\mathrm{2}^{{p}−\mathrm{1}} \geqslant{p}\Rightarrow \\ $$$$\frac{{a}^{{p}} +\mathrm{1}}{{a}+\mathrm{1}}\:\geqslant{p}\:+\mathrm{1}>{p}\: \\ $$$$ \\ $$$$ \\ $$

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