Prove-or-disprove-that-there-is-a-positive-integer-suitable-for-n-3-1-n-n-is-divided-by-n-3-1-n-Z-

Question Number 60808 by naka3546 last updated on 26/May/19

P r o v e o r d i s p r o v e t h a t t h e r e i s

a p o s i t i v e i n t e g e r s u i t a b l e f o r

n^{3} + 1 ∣ n! (n! i s d i v i d e d b y n^{3} + 1)

n \in Z^{+}

Commented by mr W last updated on 26/May/19

t h e f i r s t o n e i s n = 17 :

n! = 355687248096000

n^{3} + 1 = 4914

n! m o d (n^{3} + 1) = 0

Commented by mr W last updated on 26/May/19

i f o u n d t h e n e x t o n e : n = 31

n! = 8222838654177922817725562880000000

n^{3} + 1 = 29792

n! m o d (n^{3} + 1) = 0

Commented by mr W last updated on 26/May/19

t h e n e x t o n e i s n = 32

Commented by naka3546 last updated on 26/May/19

h o w t o g e t t h e m, s i r ?

Commented by mr W last updated on 26/May/19

t h e r e i s n o i n t e l l i g e n t w a y b u t t r y i n g

o n e a f t e r o n e .

Commented by MJS last updated on 27/May/19

n = 32 is wrong

I get

n = 0

n = 17

n = 31

n = 50

n = 68

n = 69

n = 75

n = 80

n = 101

n = 103

n = 122

n = 147

n = 155

n = 159

n = 160

n = 164

n = 170

n = 173

n = 179

n = 182

for 0 ⩽ n ⩽ 200

\dots the next one is 212

Commented by MJS last updated on 27/May/19

btw

(n^{3} + 1) ∣ n! \Leftrightarrow (n^{3} - 1) ∣ (n - 1)!

Commented by mr W last updated on 28/May/19

t h a n k s s i r!

y o u a r e r i g h t, 32 i s w r o n g .

Answered by Rasheed.Sindhi last updated on 28/May/19

T r y i n g t o m a k e s e a r c h n a r r o w e r

n^{3} + 1 ∣ n!

(n + 1) (n^{2} - n + 1) ∣ n!

n + 1 ∣ n! \land n^{2} - n + 1 ∣ n! [a b ∣ c \Rightarrow a ∣ c & b ∣ c]

∙ n + 1 ∣ n! \Rightarrow n + 1 \notin P

[A n y p r i m e > n {c a n}^{'} t d i v i d e n!]

n \neq 2 - 1, 3 - 1, 5 - 1, \dots \dots .

n \neq 1, 2, 4, 6, 10, 12, 16, 18, \dots \dots

n m a y b e 3, 5, 7, 8, 9, 11, 13, 14, 15, 17

∙ n^{2} - n + 1 ∣ n!

c 1 : n^{2} - n + 1 < n

n^{2} - 2 n + 1 < 0

{(n - 1)}^{2} < 0 I m p o s s i b l e

c 2 : n^{2} - n + 1 = n

n = 1 {d o e s n}^{'} t s a t i s f y n^{3} + 1 ∣ n!

c 3 : n^{2} - n + 1 > n

{(n - 1)}^{2} > 0 t h e o n l y c o n s i d e r a b l e c a s e

n^{2} - n + 1 > n \land n^{2} - n + 1 ∣ n!

\Rightarrow n^{2} - n + 1 \notin P

S o n + 1 & n^{2} - n + 1 b o t h a r e n o t p r i m e

H e n c e s u c h v a l u e s o f n f o r w h i c h t h e s e

t w o e x p r e s s i o n s a r e p r i m e v a l u e m u s t

b e e x c l u d e d f r o m c o n s i d e r i n g .

(S a m p l e)

| \begin{matrix} n & n + 1 & n^{2} - n + 1 & r e m \\ 1 & 2^{*} & 1 & \times \\ 2 & 3^{*} & 3^{*} & \times \\ 3 & 4 & 7^{*} & \times \\ 4 & 5^{*} & 13^{*} & \times \\ 5 & 6 & 21 & - \\ 6 & 7^{*} & 31^{*} & \times \\ 7 & 8 & 43^{*} & \times \\ 8 & 9 & 57 & - \\ 9 & 10 & 73^{*} & \times \\ 10 & 11^{*} & 91 & \times \\ 11 & 12 & 121 & - \\ 12 & 13^{*} & 133 & \times \end{matrix} |

\times m e a n s e x c l u d e d

- m e a n s c o n s i d e r a b l e

n u m b e r s s t a r r e d a r e p r i m e .

Commented by Rasheed.Sindhi last updated on 28/May/19

CONCLUSION :

T R Y o n l y t h o s e v a l u e s f o r n,

f o r w h i c h n & n (n - 1) a r e n o t

o f p - 1 t y p e . (p i s p r i m e)

Commented by mr W last updated on 29/May/19

t h a n k s a l o t s i r!

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