Is-that-right-IF-k-1-n-n-k-n-1-k-2-so-n-is-a-prime-number-

Question Number 182934 by moh777 last updated on 17/Dec/22

I s t h a t r i g h t!

I F :

\underset{k = 1}{\sum^{n}} (⌊ \frac{n}{k} ⌋ - ⌊ \frac{n - 1}{k} ⌋) = 2

s o n i s a p r i m e n u m b e r .

Answered by dre23 last updated on 17/Dec/22

\Leftrightarrow \forall n ⩾ 2

\underset{k = 2}{\sum^{n}} ([\frac{n}{k}] - [\frac{n - 1}{k}]) = 1 n e s t p r e m i e r s

n = 2 w o r c k s \forall n ⩾ 3

\Leftrightarrow \underset{k = 2}{\sum^{n - 1}} ([\frac{n}{k}] - [\frac{n - 1}{k}]) = 0

f (n, k) = [\frac{n}{k}] - [\frac{n - 1}{k}]

i f n i s n o t p r i m e \exists s \in N - {1, n} s u c h n = s p

\frac{n}{s} = p a n d \frac{n - 1}{s} = p - \frac{1}{p},

[\frac{n}{k}] = p, [\frac{n - 1}{k}] = p - 1

\underset{k = 2}{\sum^{n - 1}} ([\frac{n}{k}] - [\frac{n - 1}{k}]) = \underset{k = 2, k \neq k}{\sum^{n - 1}} f (n, k) + p - (p - 1) ⩾ p - (p - 1) = 1

\Leftrightarrow \underset{k = 2}{\sum^{n - 1}} f (n, k) ⩾ 1 > 0 \Rightarrow n o t t r u e

s o \underset{k = 1}{\sum^{n}} f (n, k) = 2 \Rightarrow n i s p r i m e t r u e

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