Question Number 4199 by Yozzii last updated on 01/Jan/16
$${Let}\:{p}\in\mathbb{P}\:{and}\:{m}\in\mathbb{Z}^{+} . \\ $$$${Find}\:\left({p},{m}\right)\:{such}\:{that}\:{p}^{{m}−\mathrm{1}} \left({p}−\mathrm{1}\right)=\mathrm{146410}. \\ $$
Answered by Rasheed Soomro last updated on 01/Jan/16
$${p}^{{m}−\mathrm{1}} \left({p}−\mathrm{1}\right)=\mathrm{146410} \\ $$$$\mathrm{146410}=\mathrm{2}×\mathrm{5}×\mathrm{11}^{\mathrm{4}} \\ $$$$\:{p}\in\mathbb{P}\Rightarrow{p}=\mathrm{2}\:\mid\:{p}\:\:{is}\:{odd} \\ $$$${For}\:{p}=\mathrm{2},\:\:\mathrm{146410}\:{should}\:{be}\:{of}\:\mathrm{2}^{{n}} \:{type}\:{but}\:{it} \\ $$$${isn}'{t}. \\ $$$$\therefore\:{p}\:{is}\:{odd} \\ $$$$\therefore\:{p}−\mathrm{1}\:{is}\:{even} \\ $$$${even}\:{factors}\:{of}\:\mathrm{146410}\:{are}\:\mathrm{2},\mathrm{10},\mathrm{22} \\ $$$${p}−\mathrm{1}=\mathrm{2},\mathrm{10},\mathrm{22}\:\Rightarrow\:{p}=\mathrm{3},\mathrm{11},\mathrm{23} \\ $$$$\because\:\mathrm{3}\:{and}\:\mathrm{23}\:{are}\:{not}\:{factors} \\ $$$$\therefore\:{p}=\mathrm{11} \\ $$$$\mathrm{11}^{{m}−\mathrm{1}} \left(\mathrm{11}−\mathrm{1}\right)=\mathrm{146410} \\ $$$$\mathrm{11}^{{m}−\mathrm{1}} =\mathrm{11}^{\mathrm{4}} \\ $$$${m}−\mathrm{1}=\mathrm{4} \\ $$$${m}=\mathrm{5} \\ $$$$\left({p},{m}\right)=\left(\mathrm{11},\mathrm{5}\right) \\ $$