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Any-integer-s-which-fulfill-n-5-5n-3-5n-1-n-




Question Number 61782 by naka3546 last updated on 08/Jun/19
Any  integer(s)  which  fulfill            n^5  − 5n^3  + 5n + 1  ∣  n!   ?
$${Any}\:\:{integer}\left({s}\right)\:\:{which}\:\:{fulfill} \\ $$$$\:\:\:\:\:\:\:\:\:\:{n}^{\mathrm{5}} \:−\:\mathrm{5}{n}^{\mathrm{3}} \:+\:\mathrm{5}{n}\:+\:\mathrm{1}\:\:\mid\:\:{n}!\:\:\:? \\ $$
Commented by naka3546 last updated on 08/Jun/19
please  show  workings !
$${please}\:\:{show}\:\:{workings}\:! \\ $$
Commented by MJS last updated on 08/Jun/19
again we can only try
$$\mathrm{again}\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{try} \\ $$
Commented by MJS last updated on 08/Jun/19
n=0  no other solution for n≤300
$${n}=\mathrm{0} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{for}\:{n}\leqslant\mathrm{300} \\ $$
Commented by naka3546 last updated on 08/Jun/19
there  was  n  so  huge  that  fulfill  it .  But,  I  can′t  still  understand  until  now .  e.g :  Let   X_m (n)  =  (x^n  + (1/x^n ))^m   for  m = 5 .   Could   anyone  explain  to  get  that   equation  is ?
$${there}\:\:{was}\:\:{n}\:\:{so}\:\:{huge}\:\:{that}\:\:{fulfill}\:\:{it}\:. \\ $$$${But},\:\:{I}\:\:{can}'{t}\:\:{still}\:\:{understand}\:\:{until}\:\:{now}\:. \\ $$$${e}.{g}\::\:\:{Let}\:\:\:{X}_{{m}} \left({n}\right)\:\:=\:\:\left({x}^{{n}} \:+\:\frac{\mathrm{1}}{{x}^{{n}} }\right)^{{m}} \\ $$$${for}\:\:{m}\:=\:\mathrm{5}\:.\:\:\:{Could}\:\:\:{anyone}\:\:{explain}\:\:{to}\:\:{get}\:\:{that}\:\:\:{equation}\:\:{is}\:? \\ $$
Commented by naka3546 last updated on 08/Jun/19
or  may be  like  this     X_m (n)  =  (x^m  + (1/x^m ))^n
$${or}\:\:{may}\:{be}\:\:{like}\:\:{this} \\ $$$$\:\:\:{X}_{{m}} \left({n}\right)\:\:=\:\:\left({x}^{{m}} \:+\:\frac{\mathrm{1}}{{x}^{{m}} }\right)^{{n}} \\ $$

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