The arithmetic function $\lambda(n)=(-1)^{a_1+\cdots +a_k}$

Question

Define $\lambda(1)=1$, and if $n=p_1^{a_1}\cdots p_k^{a_k}$, define $$\lambda(n)=(-1)^{a_1+\cdots +a_k}$$

How can I see that $$\sum_{d\mid n}\lambda(d)=\begin{cases} 1 \,\,\text{ if $n$ is a square}\\ 0 \,\,\text{ otherwise} \end{cases} $$?

Answer 1

If you put $f(n)=\sum_{d|n} \lambda(d)$ and $$g(n)=\left\{ \begin{array}{c} 1 & \textrm{if $n$ is square}\\ 0 & \textrm{otherwise} \end{array} \right. $$ then you can see easily that if $n,m$ are coprime : $$ f(nm)=f(n)f(m)\\ g(nm)=g(n)g(m) $$ and $$ f(p^k)=g(p^k) \qquad \textrm{for all $p$ prime number} $$ so $f=g$

Answer 2

Define a $b$-tuple on $(a_1,\dots,a_k)$ as $[b_1,\dots,b_k]$ with $0\le b_i\le a_i,\;\forall 1\le i\le k$.

Then if $a_i$ is odd, there are an even number of possibilities for $b_i$, and if $a_i$ is even, there are an odd number of possibilities for $b_i$.

So unless every $a_i$ is even, i.e. $n$ is a square, we have an even number of possible $b$-tuples, with an equal number of odd and even sums.

These represent the divisors of $n$, and so $\sum_\limits{d|n} \lambda(n)=0$ unless $n$ is a square, when there are an excess of even $b$-tuples, and so in this case $\sum_\limits{d|n} \lambda(n)=1$.

Answer 3

With the prime factorization of $n$ being

$$n = \prod_p p^v$$

we have

$$\sum_{d|n} \lambda(d) = \prod_p \left(1 + (-1) + 1 + (-1) + \cdots + (-1)^v\right).$$

Now if $v$ is odd the corresponding factor is zero, so the sum is zero in this case as well. Therefore for this to be non-zero all $v$ must be even. This yields the value one for the sum terms, producing one as the end result. But if all $v$ are even $n$ certainly is a square as claimed.