Question Number 55914 by peter frank last updated on 06/Mar/19
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19
$$\left({k}−\mathrm{1}\right)\mid\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({k}−\mathrm{1}\right)\left(\frac{\mathrm{4}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{4}^{{k}+\mathrm{2}} =\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}=\frac{\mathrm{1}+\frac{\mathrm{1}}{{k}}}{\mathrm{1}−\frac{\mathrm{1}}{{k}}} \\ $$$${wait}... \\ $$$$\left.\mathrm{1}\right){let}\:{k}=+{ve}\:{integer}\:{then}\:{left}\:{hand}\:{side}\:{in}\:{the} \\ $$$${form}\:{of}\:\mathrm{4}^{{n}} \:\:{but}\:{right}\:{handside}\:{in}\:\frac{{p}}{{q}}\:{form}\: \\ $$$${so}\:{k}\neq+{ve}\:{integer} \\ $$$$\left.\mathrm{2}\right){let}\:{k}=−{ve}\:{integer}\:{then}\:{LHS}=\frac{\mathrm{1}}{\mathrm{4}^{{n}} } \\ $$$${RHS}=\frac{\mathrm{1}}{\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)}\:\:{that}\:{imply}\:\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)\:{is}\:{divisble}\:{by}\:\mathrm{4} \\ $$$${wait}... \\ $$