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

Prove  or  disprove  that  there  is  a  positive  integer  suitable  for         n^3  + 1   ∣   n!                  (  n!   is  divided  by  n^3  + 1  )  n  ∈  Z^+

$${Prove}\:\:{or}\:\:{disprove}\:\:{that}\:\:{there}\:\:{is} \\ $$$${a}\:\:{positive}\:\:{integer}\:\:{suitable}\:\:{for} \\ $$$$\:\:\:\:\:\:\:{n}^{\mathrm{3}} \:+\:\mathrm{1}\:\:\:\mid\:\:\:{n}!\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:{n}!\:\:\:{is}\:\:{divided}\:\:{by}\:\:{n}^{\mathrm{3}} \:+\:\mathrm{1}\:\:\right) \\ $$$${n}\:\:\in\:\:\mathbb{Z}^{+} \\ $$

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

the first one is n=17:  n!=355687248096000  n^3 +1=4914  n! mod (n^3 +1)=0

$${the}\:{first}\:{one}\:{is}\:{n}=\mathrm{17}: \\ $$$${n}!=\mathrm{355687248096000} \\ $$$${n}^{\mathrm{3}} +\mathrm{1}=\mathrm{4914} \\ $$$${n}!\:{mod}\:\left({n}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$

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

i found the next one: n=31  n!=8222838654177922817725562880000000  n^3 +1=29792  n! mod (n^3 +1)=0

$${i}\:{found}\:{the}\:{next}\:{one}:\:{n}=\mathrm{31} \\ $$$${n}!=\mathrm{8222838654177922817725562880000000} \\ $$$${n}^{\mathrm{3}} +\mathrm{1}=\mathrm{29792} \\ $$$${n}!\:{mod}\:\left({n}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$

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

the next one is n=32

$${the}\:{next}\:{one}\:{is}\:{n}=\mathrm{32} \\ $$

Commented by naka3546 last updated on 26/May/19

how  to  get  them,  sir?

$${how}\:\:{to}\:\:{get}\:\:{them},\:\:{sir}? \\ $$

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

there is no intelligent way but trying  one after one.

$${there}\:{is}\:{no}\:{intelligent}\:{way}\:{but}\:{trying} \\ $$$${one}\:{after}\:{one}. \\ $$

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  ...the next one is 212

$${n}=\mathrm{32}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$${n}=\mathrm{0} \\ $$$${n}=\mathrm{17} \\ $$$${n}=\mathrm{31} \\ $$$${n}=\mathrm{50} \\ $$$${n}=\mathrm{68} \\ $$$${n}=\mathrm{69} \\ $$$${n}=\mathrm{75} \\ $$$${n}=\mathrm{80} \\ $$$${n}=\mathrm{101} \\ $$$${n}=\mathrm{103} \\ $$$${n}=\mathrm{122} \\ $$$${n}=\mathrm{147} \\ $$$${n}=\mathrm{155} \\ $$$${n}=\mathrm{159} \\ $$$${n}=\mathrm{160} \\ $$$${n}=\mathrm{164} \\ $$$${n}=\mathrm{170} \\ $$$${n}=\mathrm{173} \\ $$$${n}=\mathrm{179} \\ $$$${n}=\mathrm{182} \\ $$$$\mathrm{for}\:\mathrm{0}\leqslant{n}\leqslant\mathrm{200} \\ $$$$...\mathrm{the}\:\mathrm{next}\:\mathrm{one}\:\mathrm{is}\:\mathrm{212} \\ $$

Commented by MJS last updated on 27/May/19

btw  (n^3 +1)∣n! ⇔ (n^3 −1)∣(n−1)!

$$\mathrm{btw} \\ $$$$\left({n}^{\mathrm{3}} +\mathrm{1}\right)\mid{n}!\:\Leftrightarrow\:\left({n}^{\mathrm{3}} −\mathrm{1}\right)\mid\left({n}−\mathrm{1}\right)! \\ $$

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

thanks sir!  you are right, 32 is wrong.

$${thanks}\:{sir}! \\ $$$${you}\:{are}\:{right},\:\mathrm{32}\:{is}\:{wrong}. \\ $$

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

Trying to make search narrower         n^3  + 1   ∣   n!             (n+1)(n^2 −n+1)   ∣   n!      n+1 ∣ n!   ∧  n^2 −n+1 ∣  n!  [ ab ∣ c ⇒a∣c & b∣c ]      •  n+1  ∣  n! ⇒ n+1 ∉  P       [ Any prime>n can′t divide n! ]  n≠2−1,3−1,5−1,.......  n≠1,2,4,6,10,12,16,18,......  n may be 3,5,7,8,9,11,13,14,15,17     •  n^2 −n+1 ∣  n!        c1:n^2 −n+1<n              n^2 −2n+1<0              (n−1)^2 <0     Impossible       c2: n^2 −n+1=n             n=1  doesn′t satisfy n^3 +1 ∣ n!       c3: n^2 −n+1>n               (n−1)^2 >0    the only considerable case       n^2 −n+1>n  ∧  n^2 −n+1 ∣ n!        ⇒  n^2 −n+1 ∉ P   So n+1 &  n^2 −n+1 both are not prime  Hence such values of n for which these  two expressions are prime value must  be excluded from considering.  (Sample)   determinant ((n,(n+1),(n^2 −n+1),(rem)),(1,(   2^∗ ),(         1),(   ×)),(2,(   3^∗ ),(         3^∗ ),(   ×)),(3,(   4),(         7^∗ ),(   ×)),(4,(   5^∗ ),(        13^∗ ),(   ×)),(5,(   6),(        21),(   −)),(6,(   7^∗ ),(        31^∗ ),(   ×)),(7,(   8),(        43^∗ ),(   ×)),(8,(    9),(       57),(   −)),(9,(   10),(      73^∗ ),(   ×)),((10),(   11^∗ ),(      91),(   ×)),((11),(   12),(     121),(   −)),((12),(   13^∗ ),(     133),(    ×)))  × means excluded  − means considerable  numbers starred are prime.

$${Trying}\:{to}\:{make}\:{search}\:{narrower} \\ $$$$\:\:\:\:\:\:\:{n}^{\mathrm{3}} \:+\:\mathrm{1}\:\:\:\mid\:\:\:{n}!\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)\:\:\:\mid\:\:\:{n}! \\ $$$$\:\:\:\:{n}+\mathrm{1}\:\mid\:{n}!\:\:\:\wedge\:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}\:\mid\:\:{n}!\:\:\left[\:{ab}\:\mid\:{c}\:\Rightarrow{a}\mid{c}\:\&\:{b}\mid{c}\:\right]\: \\ $$$$ \\ $$$$\:\bullet\:\:{n}+\mathrm{1}\:\:\mid\:\:{n}!\:\Rightarrow\:{n}+\mathrm{1}\:\notin\:\:\mathbb{P} \\ $$$$\:\:\:\:\:\left[\:{Any}\:{prime}>{n}\:{can}'{t}\:{divide}\:{n}!\:\right] \\ $$$${n}\neq\mathrm{2}−\mathrm{1},\mathrm{3}−\mathrm{1},\mathrm{5}−\mathrm{1},....... \\ $$$${n}\neq\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{10},\mathrm{12},\mathrm{16},\mathrm{18},...... \\ $$$${n}\:{may}\:{be}\:\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{11},\mathrm{13},\mathrm{14},\mathrm{15},\mathrm{17} \\ $$$$ \\ $$$$\:\bullet\:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}\:\mid\:\:{n}! \\ $$$$\:\:\:\:\:\:{c}\mathrm{1}:{n}^{\mathrm{2}} −{n}+\mathrm{1}<{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left({n}−\mathrm{1}\right)^{\mathrm{2}} <\mathrm{0}\:\:\:\:\:{Impossible} \\ $$$$\:\:\:\:\:{c}\mathrm{2}:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}={n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{n}=\mathrm{1}\:\:{doesn}'{t}\:{satisfy}\:{n}^{\mathrm{3}} +\mathrm{1}\:\mid\:{n}! \\ $$$$\:\:\:\:\:{c}\mathrm{3}:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}>{n}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left({n}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0}\:\:\:\:{the}\:{only}\:{considerable}\:{case} \\ $$$$\:\:\:\:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}>{n}\:\:\wedge\:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}\:\mid\:{n}! \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}\:\notin\:\mathbb{P} \\ $$$$\:{So}\:{n}+\mathrm{1}\:\&\:\:{n}^{\mathrm{2}} −{n}+\mathrm{1}\:{both}\:{are}\:{not}\:{prime} \\ $$$${Hence}\:{such}\:{values}\:{of}\:{n}\:{for}\:{which}\:{these} \\ $$$${two}\:{expressions}\:{are}\:{prime}\:{value}\:{must} \\ $$$${be}\:{excluded}\:{from}\:{considering}. \\ $$$$\left(\mathcal{S}{ample}\right) \\ $$$$\begin{vmatrix}{{n}}&{{n}+\mathrm{1}}&{{n}^{\mathrm{2}} −{n}+\mathrm{1}}&{{rem}}\\{\mathrm{1}}&{\:\:\:\mathrm{2}^{\ast} }&{\:\:\:\:\:\:\:\:\:\mathrm{1}}&{\:\:\:×}\\{\mathrm{2}}&{\:\:\:\mathrm{3}^{\ast} }&{\:\:\:\:\:\:\:\:\:\mathrm{3}^{\ast} }&{\:\:\:×}\\{\mathrm{3}}&{\:\:\:\mathrm{4}}&{\:\:\:\:\:\:\:\:\:\mathrm{7}^{\ast} }&{\:\:\:×}\\{\mathrm{4}}&{\:\:\:\mathrm{5}^{\ast} }&{\:\:\:\:\:\:\:\:\mathrm{13}^{\ast} }&{\:\:\:×}\\{\mathrm{5}}&{\:\:\:\mathrm{6}}&{\:\:\:\:\:\:\:\:\mathrm{21}}&{\:\:\:−}\\{\mathrm{6}}&{\:\:\:\mathrm{7}^{\ast} }&{\:\:\:\:\:\:\:\:\mathrm{31}^{\ast} }&{\:\:\:×}\\{\mathrm{7}}&{\:\:\:\mathrm{8}}&{\:\:\:\:\:\:\:\:\mathrm{43}^{\ast} }&{\:\:\:×}\\{\mathrm{8}}&{\:\:\:\:\mathrm{9}}&{\:\:\:\:\:\:\:\mathrm{57}}&{\:\:\:−}\\{\mathrm{9}}&{\:\:\:\mathrm{10}}&{\:\:\:\:\:\:\mathrm{73}^{\ast} }&{\:\:\:×}\\{\mathrm{10}}&{\:\:\:\mathrm{11}^{\ast} }&{\:\:\:\:\:\:\mathrm{91}}&{\:\:\:×}\\{\mathrm{11}}&{\:\:\:\mathrm{12}}&{\:\:\:\:\:\mathrm{121}}&{\:\:\:−}\\{\mathrm{12}}&{\:\:\:\mathrm{13}^{\ast} }&{\:\:\:\:\:\mathrm{133}}&{\:\:\:\:×}\end{vmatrix} \\ $$$$×\:{means}\:{excluded} \\ $$$$−\:{means}\:{considerable} \\ $$$${numbers}\:{starred}\:{are}\:{prime}. \\ $$

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

CONCLUSION:  TRY only those values for n,  for which n & n(n−1) are not  of  p−1 type. (p is prime)

$$\mathcal{CONCLUSION}: \\ $$$${TRY}\:{only}\:{those}\:{values}\:{for}\:{n}, \\ $$$${for}\:{which}\:{n}\:\&\:{n}\left({n}−\mathrm{1}\right)\:{are}\:{not} \\ $$$${of}\:\:{p}−\mathrm{1}\:{type}.\:\left({p}\:{is}\:{prime}\right) \\ $$$$ \\ $$

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

thanks alot sir!

$${thanks}\:{alot}\:{sir}! \\ $$

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