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Question Number 216841 by hardmath last updated on 22/Feb/25

$$\mathrm{Prove}\mathrm{that}:\delta \left(\mathrm{n}\right)=\sum _{\frac{\mathbf{d}}{\mathbf{n}}}\mathit{\phi }\left(\mathrm{d}\right)\mathit{\tau }\left(\frac{\mathrm{n}}{\mathrm{d}}\right)$$$$\mathit{\delta }\left(\mathrm{n}\right)=\sum _{\frac{\mathbf{d}}{\mathbf{n}}}\mathrm{d},\mathit{\tau }\left(\mathrm{n}\right)=\sum _{\frac{\mathbf{d}}{\mathbf{n}}}l\mathrm{and}\phi -\mathrm{Eyler}.\mathrm{f}$$

Answered by MrGaster last updated on 23/Feb/25

$$\mathrm{Let}f\left(n\right)=\sum _{d\mid n}\phi \left(d\right)\tau \left(\frac{n}{d}\right)$$$$\Rightarrow ,f\left(n\right)=\sum _{d\mid n}\phi \left(d\right)\sum _{k\mid \frac{n}{d}}1=\sum _{d\mid n}\phi \left(d\right)\sum _{k\mid \frac{n}{d}}1=\sum _{d\mid n}\phi \left(d\right)\left(\frac{n}{d}\right)$$$$\sum _{d\mid n}\phi \left(d\right)=n$$$$\mathrm{have}:f\left(n\right)=\sum _{d\mid n}\phi \left(d\right)\left(\frac{n}{d}\right)=n\sum _{d\mid n}\frac{\phi \left(d\right)}{d}$$$$\mathrm{But}\sum _{d\mid n}=\frac{\phi \left(d\right)}{d}=1$$$$\Rightarrow f\left(n\right)=n\times 1=n$$$$\therefore \delta \left(n\right)=n$$$$$$

Commented by hardmath last updated on 23/Feb/25

$$$$Excellent solution, thank you very much dear professor

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