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

Commented by MJS last updated on 01/Feb/20

$are there prime un - integers ?! ?$
Commented by mind is power last updated on 02/Feb/20
$p must bee prime ≥3 in this case let P={a∈N ∣ a is prime} a^p +1=(a+1)(Σ_(k=0) ^(n−1) (−1)^(n−k) a^k ) if a=1⇒((a^p +1)/(a+1))=1∉P if a≥2 ⇒ ((a^p +1)/(a+1))=a^(p−1) −a^(p−2) +.........+1≥a^(p−1) +a^(p−3) +........+1 claim p(k) 2^(k−1) ≥k ∀k∈N^∗ for k=1 2^0 ≥1 true Equslity ,∀k∈N assum p(k) true 2^k =2.2^(k−1) ≥2(k)=2k ⇒2^k ≥2k=k+k k≥1⇒2^k ≥k+1 P(k)⇒p(k+1) since P(k) true ⇒∀k∈N^∗ 2^(k−1) ≥k a^(p−1) ≥2^(p−1) ≥p⇒ ((a^p +1)/(a+1)) ≥p +1>p$
$p m u s t b e e p r i m e ⩾ 3$ $i n t h i s c a s e$ $l e t P = {a \in N ∣ a i s p r i m e}$ $a^{p} + 1 = (a + 1) (\underset{k = 0}{\sum^{n - 1}} {(- 1)}^{n - k} a^{k})$ $i f a = 1 \Rightarrow \frac{a^{p} + 1}{a + 1} = 1 \notin P$ $i f a ⩾ 2 \Rightarrow$ $\frac{a^{p} + 1}{a + 1} = a^{p - 1} - a^{p - 2} + . . . . . . . . . + 1 ⩾ a^{p - 1} + a^{p - 3} + . . . . . . . . + 1$ $c l a i m p (k) 2^{k - 1} ⩾ k \forall k \in N^{}$ $f o r k = 1 2^{0} ⩾ 1 t r u e E q u s l i t y, \forall k \in N a s s u m p (k) t r u e$ $2^{k} = 2 . 2^{k - 1} ⩾ 2 (k) = 2 k$ $\Rightarrow 2^{k} ⩾ 2 k = k + k$ $k ⩾ 1 \Rightarrow 2^{k} ⩾ k + 1$ $P (k) \Rightarrow p (k + 1) s i n c e P (k) t r u e \Rightarrow \forall k \in N^{} 2^{k - 1} ⩾ k$ $a^{p - 1} ⩾ 2^{p - 1} ⩾ p \Rightarrow$ $\frac{a^{p} + 1}{a + 1} ⩾ p + 1 > p$
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