Liouville lambda function

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The Liouville lambda function $\lambda (n)=(-1)^{\Omega (n)}$, where $\Omega (n)$ is the number of prime factors function (sometimes called "big omega"), tells whether $n$ has an odd or even number of prime factors, not necessarily distinct. The Liouville lambda function is sometimes denoted by $\psi (n)$. The function is completely multiplicative, meaning that $\lambda (mn)=\lambda (m)\lambda (n)$ regardless of whether $\gcd(m,n)=1$ or not.

For example, $\lambda (26)=1$ because 26 has two prime factors (2 and 13) and so $\lambda (26)=(-1)^{2}=1$. $\lambda (27)=-1$ because 27 has three prime factors (3 thrice) and so $\lambda (27)=(-1)^{3}=-1$. With $\lambda (28)=-1$ we exhibit that the function is completely multiplicative, since $\lambda (28)=\lambda (2)\lambda (14)=\lambda (4)\lambda (7)=-1\times 1=-1$. See A008836 for more values.

Theorem. The Liouville lambda function is completely multiplicative. Given $\{m,n\}\in \mathbb {Z} ^{+}$, the equality $\lambda (mn)=\lambda (m)\lambda (n)$ holds even when $\gcd(m,n)>1$.

Proof. Remember that $\Omega (mn)=\Omega (m)+\Omega (n)$ whether $\gcd(m,n)=1$ or not (whereas with the number of distinct prime factors function this would not be the case). Therefore, if $\Omega (m)$ is even and $\Omega (n)$ is also even, then so is $\Omega (mn)$ and thus $\lambda (mn)=\lambda (m)\lambda (n)=(-1)^{\Omega (m)}(-1)^{\Omega (n)}=1\times 1=1$. If $\Omega (m)$ is even but $\Omega (n)$ is odd, then $\Omega (mn)$ is odd and thus $\lambda (mn)=1\times -1=1$. But if both $\Omega (m)$ and $\Omega (n)$ are odd, then $\Omega (mn)$ is even and thus $\lambda (mn)=-1\times -1=1$. ENDOFPROOFMARK

As the Mertens function is to the Möbius function, so is the Liouville summatory function $L(n)=\sum _{i=1}^{n}\lambda (i)$ (see A002819). See Pólya's conjecture.