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Let $n \in \mathbb{Z}$ with $n > 0$. Let $F(n) = \sum_{d \mid n} \lambda(d)$. Prove that $$F(n) = \begin{cases}1, \quad \text{if }n \text{ is a perfect square}\\ 0, \quad \text{otherwise} \end{cases} $$

By the Fundamental Theorem of Arithmetic, all $d$'s admit a prime factorization $d=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ for primes $p_i$ and nonnegative integers $a_i$. So $\lambda(d)=\lambda(p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k})$. Now the idea is that all the divisors will cancel in pairs of $1$ and $-1$ when $n$ is a perfect square, except the divisor $1$, and so the sum will total $1$. How do I prove this rigorously? If I just choose a generic divisor and write it's prime factorization I don't find anything I can generalize. Can you help?

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    $\begingroup$ note that Liouville function is multiplicative so it is enough to check it for prime powers only. $\endgroup$ – leshik Oct 6 '13 at 2:45
  • $\begingroup$ So prime powers $p_i^{a_i}$ for each prime? But we have divisors which have mixed primes, like $p_1^{a_1}p_2^{a_2}$ is a divisor. I know I can split it up like $\lambda(p_1^{a_1}p_2^{a_2})=\lambda(p_1^{a_1}) \lambda(p_2^{a_2})$ but how does the argument go for the pairwise cancellation with this? $\endgroup$ – Numbersandsoon Oct 6 '13 at 2:48
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    $\begingroup$ @BoSchmidt No, what leshik means is that if $\lambda(n)$ is multiplicative then so is $\sum_{d\mid n} \lambda(d)$. Whenever you have two functions $f(n)$ and $g(n)$ which are known to be multiplicative, proving $f(p^k) = g(p^k)$ for every prime power $p^k$ is enough to show $f(n) = g(n)$ for all $n$. This avoids having to worry about mixed prime divisors. $\endgroup$ – Erick Wong Oct 6 '13 at 7:06
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    $\begingroup$ This is the Theorem 2.19 in the book "Introduction to Analytic Number Theory" of Tom M. Apostol. Springer-Verlag, New York-Heidelberg-Berlin. (En traducción al Español, está en la página 46). The theorem add the property $\lambda^{-1}(n)=|\mu (n)|$ where $\mu$ is the Möbius function. $\endgroup$ – Piquito Jul 20 '16 at 12:36
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You can prove it using "multiplicative" property and it's simple but I just solved it without it and I believe that it's much more beautiful.

Consider $$n=p_{1}^{a_1} p_2^{a_2} ... p_k^{a_k}$$

So $\Omega(n)=\sum a_i$ and $\lambda(n)=(-1)^{\Omega(n)}$

Now $$\sum_{d|n}\lambda(d)=\sum_{d|n}(-1)^{\Omega(d)}$$

So it's suffices to calculate $D$; the difference of the number of divisor with even $\Omega$ and odd $\Omega.$

Now consider $f(x)=(1+x+x^2+...+x^{a_1})(1+x+x^2+...+x^{a_2})...(1+x+x^2+...+x^{a_k}) $

The coefficient of $x^r$ in above expansion is equal to the number of solutions of this equation:

$$x_1+x_2+...+x_k=r $$ $$0 \le x_i \le a_i$$ which is the number of divisors of $n$ with $\Omega$ equals to $r$.

Hence, $f(-1)=D.$

But $f(-1)$ is $0$ if at least one of $a_i$ is odd and $f(-1)=1$ if $n$ is a perfect square. $Q.E.D$


As a second solution:

It's easy to check $\lambda=1_{Sq}*\mu $

$[$ Actually the summation has only one term.$]$

Now convolve both sides with the $1$ function [which is inverse of $\mu$]

$$ \lambda * 1=1_{Sq}*\mu*1=1_{Sq} $$

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    $\begingroup$ the most natural answer is that $\zeta(s) = \prod_p \frac{1}{1-p^{-s}}$ so that $\frac{\zeta(2s)}{\zeta(s)} = \prod_p \frac{1}{1+p^{-s}}= \sum_n \lambda(n) n^{-s}$ and $\sum_n \sum_{d | n} \lambda(d) n^{-s} = (\sum_n \lambda(n) n^{-s})(\sum_n n^{-s})$ $ = \frac{\zeta(2s)}{\zeta(s)} \zeta(s) = \zeta(2s) = \sum_n n^{-2s} = \sum_n n^{-s} 1_{n \text{ is a square}}$ $\endgroup$ – reuns Jul 20 '16 at 15:21
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    $\begingroup$ @user1952009 it was exactly what i said in the second proof... $\endgroup$ – MR_BD Jul 20 '16 at 15:32
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    $\begingroup$ no because what I wrote is clear and rigorous :) $\endgroup$ – reuns Jul 20 '16 at 15:33
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    $\begingroup$ @user1952009 it's what you think yourself! :D $\endgroup$ – MR_BD Jul 20 '16 at 15:36
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    $\begingroup$ @user1952009 Convolution is the base of Dirichlet series production and is as powerful as Dirichlet series $\endgroup$ – MR_BD Jul 20 '16 at 15:41

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