Question Number 69064 by Sayantan chakraborty last updated on 18/Sep/19
Commented by Sayantan chakraborty last updated on 18/Sep/19
$$\mathrm{PLEASE}\:\mathrm{HELP} \\ $$
Commented by Sayantan chakraborty last updated on 19/Sep/19
$$\mathrm{please}\:\mathrm{solve}\:\mathrm{it}. \\ $$
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 [?]](Q69102.png)
$${y}=\left(\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{{n}} \\ $$$${y}\left\{{y}\right\}\notin\mathbb{Z}\forall{n}=\mathrm{2}{k};\:{k}\in\mathbb{N}^{\bigstar} \:\left[?\left[\right.\right. \\ $$$${y}\left\{{y}\right\}=\mathrm{5}^{{n}} \in\mathbb{Z}\forall{n}=\mathrm{2}{k}−\mathrm{1};\:{k}\in\mathbb{N}^{\bigstar} \:\left[?\right] \\ $$$$\Rightarrow \\ $$$${f}\left({k}\right)=\mathrm{5}^{\mathrm{2}{k}−\mathrm{1}} +\mathrm{25}^{\mathrm{2}{k}−\mathrm{1}} +\mathrm{125}^{\mathrm{2}{k}−\mathrm{1}} \\ $$$${x}={f}\left(\mathrm{45}\right)=\mathrm{5}^{\mathrm{89}} +\mathrm{25}^{\mathrm{89}} +\mathrm{125}^{\mathrm{89}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{show}\:\mathrm{for}\:\mathrm{which}\:\mathrm{values}\:\mathrm{of}\:{k} \\ $$$$\mathrm{31}\mid\left(\mathrm{5}^{\mathrm{2}{k}−\mathrm{1}} +\mathrm{25}^{\mathrm{2}{k}−\mathrm{1}} +\mathrm{125}^{\mathrm{2}{k}−\mathrm{1}} \right)\:\left[?\right] \\ $$$$ \\ $$$$\mathrm{trying}... \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}\:\mathrm{for}\:{k}\in\left\{\mathrm{2},\:\mathrm{5},\:\mathrm{8},\:\mathrm{11},\:\mathrm{14},\:\mathrm{17},\:\mathrm{20},\:...\right\} \\ $$$$\mathrm{looks}\:\mathrm{like}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}\:\mathrm{for}\:{k}=\mathrm{2}+\mathrm{3}{l};\:{l}\in\mathbb{N}\:\left[?\right] \\ $$$${n}=\mathrm{89}\:\Leftrightarrow\:{k}=\mathrm{45}\:\Leftrightarrow\:{k}\neq\mathrm{2}+\mathrm{3}{l}\forall{l}\in\mathbb{N} \\ $$$$\Rightarrow\:\mathrm{31}\mid\left({x}\left\{{x}\right\}+{x}^{\mathrm{2}} \left\{{x}\right\}^{\mathrm{2}} +{x}^{\mathrm{3}} \left\{{x}\right\}^{\mathrm{3}} \right);\:{x}=\left(\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{\mathrm{89}} \\ $$$$ \\ $$$$\mathrm{needed}\:\mathrm{proofs}\:\mathrm{are}\:\mathrm{marked}\:\mathrm{with}\:\left[?\right] \\ $$
Answered by mind is power last updated on 19/Sep/19
$${i}\:{will}\:{try}\:\:\:{is}\:{not}\:{easy}\:! \\ $$
Commented by Sayantan chakraborty last updated on 20/Sep/19
$$\mathrm{It}\:\mathrm{is}\:\mathrm{challenging}\:\mathrm{problem} \\ $$