It is about a day i am working on this question with no success. $f(n) = [\sqrt{n}] - [\sqrt{n-1}]$ is not completely multiplicative example is n=20, m=80. [x] is integer part of x. Thanks in advance for reading and helping out.
for f to be multiplicative, gcd(m,n)=1 is must.
Note that $$f(n)=\begin{cases}1&\text{if $n$ is a perfect square}\\0&\text{otherwise}\end{cases} $$ This is so because if $n=k^2$, then $\lfloor\sqrt n\rfloor =k$ and $\lfloor \sqrt{n-1}\rfloor=k-1$, wheras for $n$ with $k^2<n<(k+1)^2$, we have $\lfloor\sqrt n\rfloor =k=\lfloor \sqrt{n-1}\rfloor$.
Now assume $n,m$ are coprime. Then $nm$ is a perfect square iff both $n$ and $m$ are. Hence $f$ is multiplicative.