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

Let σ(n) be the sum of all positive divisors of the integer n and let p be any prime number. show that σ(n)<2n holds true for all n of the form n=p^2 . Mastermind

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

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

σ(p^2 )=1+p+p^2 <2p^2 ⇔p^2 >p+1 ⇔p(p−1)>1, since p≥2⇒p−1≥1 ⇒p(p−1)≥p>1

$$\:\sigma\left(\mathrm{p}^{\mathrm{2}} \right)=\mathrm{1}+\mathrm{p}+\mathrm{p}^{\mathrm{2}} <\mathrm{2p}^{\mathrm{2}} \Leftrightarrow\mathrm{p}^{\mathrm{2}} >\mathrm{p}+\mathrm{1} \\ $$ $$\:\Leftrightarrow\mathrm{p}\left(\mathrm{p}−\mathrm{1}\right)>\mathrm{1},\:\mathrm{since}\:\mathrm{p}\geqslant\mathrm{2}\Rightarrow\mathrm{p}−\mathrm{1}\geqslant\mathrm{1} \\ $$ $$\Rightarrow\mathrm{p}\left(\mathrm{p}−\mathrm{1}\right)\geqslant\mathrm{p}>\mathrm{1} \\ $$ $$ \\ $$

Commented byMastermind last updated on 05/Aug/22

Thanks

$$\mathrm{Thanks} \\ $$

Commented byMastermind last updated on 06/Aug/22

But why did you use 1+p+p^2 ?

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

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