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I was studying here and I stumbled upon 2 problems. On the first one I just can't see the pattern between the numbers, but I've found that for all primes except 3 the equality doesn't work. The second one I don't have a clue about how to start, just giving me this hint should do it.

1) Consider the function $\Phi$ the Euler's totient function and let $\tau: \mathbb{N}\rightarrow\mathbb{N}$ be $\tau(n)= \displaystyle\sum_{d|n}1$. Determine all the integers $n$ for which $\Phi(n) = \tau(n)$.

2) If $n \in \mathbb{Z}_{+}^{*}$, then $\displaystyle\sum_{k=1}^{2n} \tau(k) - \sum_{k=1}^{n} \lfloor\frac{2n}{k}\rfloor = n$.

Thanks for the help!

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  • $\begingroup$ $$\prod_{p^k \|n} \frac{p^{k-1} (p-1)}{k+1} = ?$$ $\endgroup$ – reuns Sep 10 at 1:03
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    $\begingroup$ The term "$p^k || n$" is "$p^k$ divides $n$"? Just for notation sake. $\endgroup$ – user447599 Sep 10 at 1:32
  • $\begingroup$ $\tau(n) = \prod_{p^k \| n} \tau(p^k)$ means it is a multiplicative function $\endgroup$ – reuns Sep 10 at 1:35
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Hint: For the second one, notice that $\lfloor \frac{2n}{k}\rfloor=|\{i\leq 2n:k|i\}|$ and $\sum _{k=1}^{2n}\tau (k)=\sum _{i=1}^{2n}|\{k\leq 2n:i|k\}|.$ Also, notice that the limits in the sum are different, so in between $n+1$ and $2n$ you are missing a $1$[why?].

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