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Question Number 174594 by Mastermind last updated on 05/Aug/22

$$\mathrm{Let}\sigma \left(\mathrm{n}\right)\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{positive}$$$$\mathrm{divisors}\mathrm{of}\mathrm{the}\mathrm{integer}\mathrm{n}\mathrm{and}\mathrm{let}\mathrm{p}\mathrm{be}$$$$\mathrm{any}\mathrm{prime}\mathrm{number}.\mathrm{show}\mathrm{that}$$$$\sigma \left(\mathrm{n}\right)<2\mathrm{n}\mathrm{holds}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{n}\mathrm{of}\mathrm{the}$$$$\mathrm{form}\mathrm{n}={\mathrm{p}}^2.$$$$$$$$\mathrm{Mastermind}$$

Answered by floor(10²Eta[1]) last updated on 05/Aug/22

$$\sigma \left({\mathrm{p}}^2\right)=1+\mathrm{p}+{\mathrm{p}}^2<{2\mathrm{p}}^2\iff {\mathrm{p}}^2>\mathrm{p}+1$$$$\iff \mathrm{p}(\mathrm{p}-1)>1,\mathrm{since}\mathrm{p}⩾2\Rightarrow \mathrm{p}-1⩾1$$$$\Rightarrow \mathrm{p}(\mathrm{p}-1)⩾\mathrm{p}>1$$$$$$

Commented by Mastermind last updated on 05/Aug/22

$$\mathrm{Thanks}$$

Commented by Mastermind last updated on 06/Aug/22

$$\mathrm{But}\mathrm{why}\mathrm{did}\mathrm{you}\mathrm{use}1+\mathrm{p}+{\mathrm{p}}^2?$$

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