Subfield: Number Theory


Question: Let $\tau(n)$ be the number of positive divisors of $n$. Let $\tau_{1}(n)$ be the number of positive divisors of $n$ which have remainders 1 when divided by 3 . Find all possible integral values of the fraction $\frac{\tau(10 n)}{\tau_{1}(10 n)}$.

Solution:
In this solution, we always use $p_{i}$ to denote primes congruent to $1 \bmod 3$, and use $q_{j}$ to denote primes congruent to $2 \bmod 3$. When we express a positive integer $m$ using its prime factorization, we also include the special case $m=1$ by allowing the exponents to be zeros. We first compute $\tau_{1}(m)$ for a positive integer $m$.

- Claim. Let $m=3^{x} p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a_{s}} q_{1}^{b_{1}} q_{2}^{b_{2}} \cdots q_{t}^{b_{t}}$ be the prime factorization of $m$. Then

$$
\tau_{1}(m)=\prod_{i=1}^{s}\left(a_{i}+1\right)\left\lceil\frac{1}{2} \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rceil\tag{1}
$$

Proof. To choose a divisor of $m$ congruent to $1 \bmod 3$, it cannot have the prime divisor 3 , while there is no restriction on choosing prime factors congruent to 1 mod 3. Also, we have to choose an even number of prime factors (counted with multiplicity) congruent to $2 \bmod 3$.

If $\prod_{j=1}^{t}\left(b_{j}+1\right)$ is even, then we may assume without loss of generality $b_{1}+1$ is even. We can choose the prime factors $q_{2}, q_{3}, \ldots, q_{t}$ freely in $\prod_{j=2}^{t}\left(b_{j}+1\right)$ ways. Then the parity of the number of $q_{1}$ is uniquely determined, and hence there are $\frac{1}{2}\left(b_{1}+1\right)$ ways to choose the exponent of $q_{1}$. Hence (1) is verified in this case.

If $\prod_{j=1}^{t}\left(b_{j}+1\right)$ is odd, we use induction on $t$ to count the number of choices. When $t=1$, there are $\left\lceil\frac{b_{1}+1}{2}\right\rceil$ choices for which the exponent is even and $\left\lfloor\frac{b_{1}+1}{2}\right\rfloor$ choices for which the exponent is odd. For the inductive step, we find that there are

$$
\left\lceil\frac{1}{2} \prod_{j=1}^{t-1}\left(b_{j}+1\right)\right\rceil \cdot\left\lceil\frac{b_{t}+1}{2}\right\rceil+\left\lfloor\frac{1}{2} \prod_{j=1}^{t-1}\left(b_{j}+1\right)\right\rfloor \cdot\left\lfloor\frac{b_{t}+1}{2}\right\rfloor=\left\lceil\frac{1}{2} \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rceil
$$

choices with an even number of prime factors and hence $\left\lfloor\frac{1}{2} \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rfloor$ choices with an odd number of prime factors. Hence (1) is also true in this case.

Let $n=3^{x} 2^{y} 5^{z} p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a_{s}} q_{1}^{b_{1}} q_{2}^{b_{2}} \cdots q_{t}^{b_{t}}$. Using the well-known formula for computing the divisor function, we get

$$
\tau(10 n)=(x+1)(y+2)(z+2) \prod_{i=1}^{s}\left(a_{i}+1\right) \prod_{j=1}^{t}\left(b_{j}+1\right)\tag{2}
$$

By the Claim, we have

$$
\tau_{1}(10 n)=\prod_{i=1}^{s}\left(a_{i}+1\right)\left\lceil\frac{1}{2}(y+2)(z+2) \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rceil\tag{3}
$$



If $c=(y+2)(z+2) \prod_{j=1}^{t}\left(b_{j}+1\right)$ is even, then (2) and (3) imply

$$
\frac{\tau(10 n)}{\tau_{1}(10 n)}=2(x+1)
$$

In this case $\frac{\tau(10 n)}{\tau_{1}(10 n)}$ can be any even positive integer as $x$ runs through all nonnegative integers. If $c$ is odd, which means $y, z$ are odd and each $b_{j}$ is even, then (2) and (3) imply

$$
\frac{\tau(10 n)}{\tau_{1}(10 n)}=\frac{2(x+1) c}{c+1}\tag{4}
$$

For this to be an integer, we need $c+1$ divides $2(x+1)$ since $c$ and $c+1$ are relatively prime. Let $2(x+1)=k(c+1)$. Then $(4)$ reduces to

$$
\frac{\tau(10 n)}{\tau_{1}(10 n)}=k c=k(y+2)(z+2) \prod_{j=1}^{t}\left(b_{j}+1\right)
\tag{5}
$$

Noting that $y, z$ are odd, the integers $y+2$ and $z+2$ are at least 3 . This shows the integer in this case must be composite. On the other hand, for any odd composite number $a b$ with $a, b \geqslant 3$, we may simply take $n=3^{\frac{a b-1}{2}} \cdot 2^{a-2} \cdot 5^{b-2}$ so that $\frac{\tau(10 n)}{\tau_{1}(10 n)}=a b$ from (5).

We conclude that the fraction can be any even integer or any odd composite number. Equivalently, it can be 2 or any composite number.

Final answer: All composite numbers together with 2