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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

$$y={(7+3\sqrt{6})}^n$$$$y\left\{y\right\}\notin \mathbb{Z}\mathrm{\forall }n=2k;k\in {\mathbb{N}}^{★}[?[$$$$y\left\{y\right\}=5^n\in \mathbb{Z}\mathrm{\forall }n=2k-1;k\in {\mathbb{N}}^{★}[?]$$$$\Rightarrow$$$$f\left(k\right)=5^{2k-1}+{25}^{2k-1}+{125}^{2k-1}$$$$x=f\left(45\right)=5^{89}+{25}^{89}+{125}^{89}$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{for}\mathrm{which}\mathrm{values}\mathrm{of}k$$$$31\mid (5^{2k-1}+{25}^{2k-1}+{125}^{2k-1})[?]$$$$$$$$\mathrm{trying}...$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{true}\mathrm{for}k\in \{2,5,8,11,14,17,20,...\}$$$$\mathrm{looks}\mathrm{like}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{true}\mathrm{for}k=2+3l;l\in \mathbb{N}[?]$$$$n=89\iff k=45\iff k\ne 2+3l\mathrm{\forall }l\in \mathbb{N}$$$$\Rightarrow 31\mid (x\left\{x\right\}+x^2{\left\{x\right\}}^2+x^3{\left\{x\right\}}^3);x={(7+3\sqrt{6})}^{89}$$$$$$$$\mathrm{needed}\mathrm{proofs}\mathrm{are}\mathrm{marked}\mathrm{with}[?]$$

Answered by mind is power last updated on 19/Sep/19

$$iwilltryisnoteasy!$$

Commented by Sayantan chakraborty last updated on 20/Sep/19

$$\mathrm{It}\mathrm{is}\mathrm{challenging}\mathrm{problem}$$

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