Find all completely multipiclative functions $f:\mathbb{Z}\rightarrow \mathbb{Z}_{\geqslant 0}$ such that for any $a,b\in \mathbb{Z}$ and $b\neq 0$, there exist integers $q,r$ such that
$$a=bq+r$$
and
$$f(r)<f(b)$$

Proposed by Navid Safaei
1. **Lemma 1:**
   We need to show that \( f(0) = 0 \), \( f(1) = f(-1) = 1 \), and \( f(n) > 1 \) if \( |n| > 1 \).

   - Since \(\mathbb{N}_{\geq 0}\) is well-ordered, there exists some \( z \in \mathbb{Z} \) such that \( f(z) \) is the minimum.
   - If \( z \neq 0 \), then there exist some \( q, r \in \mathbb{Z} \) such that \( zq + r = 0 \) and \( f(r) < f(z) \), which contradicts the choice of \( z \).
   - Therefore, \( f(z) \) achieves its minimum exactly when \( z = 0 \).

   By the complete multiplicativity, we have \( f(0)f(a) = f(0) \) for all \( a \in \mathbb{Z} \). Therefore, either \( f(0) = 0 \) or \( f(a) = 1 \) for all \( a \in \mathbb{Z} \). However, if \( f(a) = 1 \) for all \( a \in \mathbb{Z} \), then \( f(z) \) achieves its minimum everywhere, which is a contradiction. As a consequence, \( f(0) = 0 \).

   We also have \( f(1)^2 = f(1) \), and as \( f(1) > f(0) \), we see \( f(1) = 1 \). By \( f(-1)^2 = f(1) \), we now deduce that \( f(-1) = 1 \).

   Now, if \( f(b) = 1 \) for some \( b \in \mathbb{Z} \), then we see that \( f(r) < f(b) \) implies \( r = 0 \). This shows that \( b \) divides all integers, and thus \( |b| = 1 \).

   As a consequence, \( f \) is an even function.

2. **Lemma 2:**
   For all \( b \in \mathbb{N} \) and any integer \( 0 \leq r < b \), we have \( f(r) < f(b) \) or \( f(b-r) < f(b) \).

   - Suppose for the sake of contradiction that there exists some pair \( (b, r) \) of non-negative integers with \( 0 \leq r < b \) and \( f(r), f(b-r) \geq f(b) \).
   - Among those pairs, choose one such that \( f(b) \) is the minimum.
   - By the assumption given by the problem, and also that \( f \) is even, there exists \( b' \in \mathbb{N}_{\geq 0} \) such that \( b' \equiv \pm r \mod b \) and \( f(b') < f(b) \).
   - Choose the smallest \( b' \) that satisfies those. By the choice of \( (b, r) \), we see that \( b' > b \).
   - By the minimality of \( b' \), we have \( f(b'-b) \geq f(b) > f(b') \). Hence \( (b', b) \) is another pair with \( f(b), f(b'-b) \geq f(b') \).
   - However, since \( f(b') < f(b) \), this contradicts the choice of \( (b, r) \). Therefore, such a pair does not exist.

3. **Lemma 3:**
   For any three distinct primes \( p_1, p_2, p_3 \), if \( \log_{p_1} f(p_1) \) is the largest among \( \log_{p_i} f(p_i) \) for \( i = 1, 2, 3 \), then \( \log_{p_2} f(p_2) = \log_{p_3} f(p_3) \).

   - Assume for the sake of contradiction that \( \log_{p_2} f(p_2) > \log_{p_3} f(p_3) \).
   - Let \( \alpha = \log_{p_2} f(p_2) \) and \( \beta = \log_{p_3} f(p_3) \).
   - Let \( k \) be a sufficiently large integer to be chosen later.
   - Let \( N \) be another sufficiently large integer to be chosen later.
   - Let \( n \) be the element in \( [N] \backslash \{1\} \) that maximizes the pair \( (-\log_n f(n), n) \) with respect to the lexicographical order.
   - If \( n \leq \sqrt{N} \), then we see that \( \log_{n^2} f(n^2) = \frac{1}{2} \log_n f(n)^2 = \log_n f(n) \) and thus \( (\log_{n^2} f(n^2), n^2) \) has an even larger lexicographical order, which is a contradiction.
   - Therefore, \( n > \sqrt{N} \). Let \( \gamma = \log_n f(n) \). Note that if \( N > p_3 \), then \( \gamma \leq \beta \).

   Now consider the equation \( p_1^k x + p_2^k y = n \). Since \( \gcd(p_1^k, p_2^k) = 1 \), by Bezout's theorem, there exist \( x_0, y_0 \in \mathbb{Z} \) such that \( p_1^k x_0 + p_2^k y_0 = n \). In general, we thus have \( x = x_0 + p_2^k t \) and \( y = y_0 - p_1^k t \) for \( t \in \mathbb{Z} \). Pick \( t \) so that \( |p_1^k x - n/2| \leq p_1^k p_2^k / 2 \). Then we have \( p_1^k x, p_2^k y \geq (n - p_1^k p_2^k) / 2 \).

   We will now show that the pair \( (n, p_1^k x) \) contradicts the previous lemma as long as we choose the parameters \( k, N \) properly. Indeed, pick \( k \) such that \( p_1^{k(\alpha - \beta)}, p_2^{k(\alpha - \beta)} \geq 4^{\beta} \), and then pick \( N \geq \max((2 p_1^k p_2^k)^2, p_3) \). Then \( n > 2 p_1^k p_2^k \) and so \( p_1^k x, p_2^k y > n / 4 \). Moreover, we have that \( f(p_1^k x) = f(p_1)^k f(x) \geq p_1^{k \alpha} x^{\gamma} \geq p_1^{k(\alpha - \beta)} (p_1^k x)^{\gamma} \geq (4 p_1^k x)^{\gamma} \geq n^{\gamma} = f(n) \). Similarly, \( f(p_2^k y) \geq f(n) \). This contradicts the previous lemma, as desired.

From this lemma, we can easily see that except for a specified prime \( p_0 \), we have \( \log_p f(p) \) is a constant. This constant should therefore be an integer, which is positive by the first lemma. Thus we see that \( f(x) = |x|^n s^{v_{p_0}(x)} \) for some positive integers \( n, s \) and some prime \( p_0 \). We now verify that any function of this form is indeed a solution. Indeed, for any \( a, b \in \mathbb{Z} \) with \( b \neq 0 \), we may find \( q, r \in \mathbb{Z} \) such that \( a = bq + r \) and \( |r| < |b| \) with \( r = 0 \) or \( v_{p_0}(r) \leq v_{p_0}(b) \), as otherwise we may instead take \( r' = r - b \) if \( r > 0 \) or \( r' = r + b \) if \( r < 0 \). With this, we now see that \( f(r) < f(b) \) holds when \( r = 0 \), and when \( r \neq 0 \) we still have
\[ f(r) = |r|^n s^{v_{p_0}(r)} < |b|^n s^{v_{p_0}(b)} = f(b) \]
as \( |r| < |b| \) and \( v_{p_0}(r) \leq v_{p_0}(b) \).

The final answer is \( \boxed{ f(x) = |x|^n s^{v_{p_0}(x)} } \) for some positive integers \( n, s \) and some prime \( p_0 \).