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Question-69064




Question Number 69064 by Sayantan chakraborty last updated on 18/Sep/19
Commented by Sayantan chakraborty last updated on 18/Sep/19
PLEASE HELP
PLEASEHELP
Commented by Sayantan chakraborty last updated on 19/Sep/19
please solve it.
pleasesolveit.
Commented by MJS last updated on 20/Sep/19
y=(7+3(√6))^n   y{y}∉Z∀n=2k; k∈N^★  [?[  y{y}=5^n ∈Z∀n=2k−1; k∈N^★  [?]  ⇒  f(k)=5^(2k−1) +25^(2k−1) +125^(2k−1)   x=f(45)=5^(89) +25^(89) +125^(89)   we have to show for which values of k  31∣(5^(2k−1) +25^(2k−1) +125^(2k−1) ) [?]    trying...  it′s not true for k∈{2, 5, 8, 11, 14, 17, 20, ...}  looks like it′s not true for k=2+3l; l∈N [?]  n=89 ⇔ k=45 ⇔ k≠2+3l∀l∈N  ⇒ 31∣(x{x}+x^2 {x}^2 +x^3 {x}^3 ); x=(7+3(√6))^(89)     needed proofs are marked with [?]
y=(7+36)ny{y}Zn=2k;kN[?[y{y}=5nZn=2k1;kN[?]f(k)=52k1+252k1+1252k1x=f(45)=589+2589+12589wehavetoshowforwhichvaluesofk31(52k1+252k1+1252k1)[?]tryingitsnottruefork{2,5,8,11,14,17,20,}lookslikeitsnottruefork=2+3l;lN[?]n=89k=45k2+3llN31(x{x}+x2{x}2+x3{x}3);x=(7+36)89neededproofsaremarkedwith[?]
Answered by mind is power last updated on 19/Sep/19
i will try   is not easy !
iwilltryisnoteasy!
Commented by Sayantan chakraborty last updated on 20/Sep/19
It is challenging problem
Itischallengingproblem

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