Sixteenth Hong Kong (China) Mathematical Olympiad

The answer is $a = 1680$.
Since $ab \equiv a(-a) = -a^2 \pmod{a+b}$, we have $a+b \mid ab$ if and only if $a+b \mid a^2$. This shows $a+b$ can be any divisor of $a^2$ greater than $a$. Also, any such divisor corresponds to a unique positive integer $b$. Therefore, $M(a)$ is the number of divisors of $a^2$ greater than $a$. Since the positive divisors of $a^2$ less than $a$ are in one-to-one correspondence to the divisors of $a^2$ greater than $a$, it remains to find $a$ which has the maximum number of positive divisors.
For any $a = p_1^{c_1} p_2^{c_2} \cdots p_s^{c_s}$ where $p_1, p_2, \ldots, p_s$ are distinct prime divisors of $a$, recall that $d(a^2) = (2c_1+1)(2c_2+1)\cdots(2c_s+1)$. WLOG we may assume $c_1 \ge c_2 \ge \cdots \ge c_s$. Since $2 \times 3 \times 5 \times 7 \times 11 = 2310 > 2013$, we have $s \le 4$.
* If $s=4$, we must have $c_3 = c_4 = 1$ since $2^2 3^2 5^2 7 = 6300 > 2013$.
- If $c_2 \ge 2$, we must have $c_1 = c_2 = 2$ since $2^3 3^2 5 \cdot 7 = 2520 > 2013$. When $a = p_1^2 p_2^2 p_3 p_4$, we have $d(a^2) = 5^2 \cdot 3^2 = 225$.
- If $c_2 = 1$, we have $c_1 \le 4$ since $2^5 3 \cdot 5 \cdot 7 = 3360 > 2013$. When $a = p_1^4 p_2 p_3 p_4$, we have $d(a^2) = 9 \cdot 3^3 = 243$.

• If $s = 3$, we must have $c_3 \le 2$ since $2^33^35^3 = 27000 > 2013$.
– If $c_3 = 2$, we must have $c_2 = 2$ since $2^33^35^2 = 5400 > 2013$. Then $c_1 \le 3$ since $2^43^25^2 = 3600 > 2013$. When $a = p_1^3 p_2^2 p_3^2$, we have $d(a^2) = 7 \cdot 5^2 = 175$.
– If $c_3 = 1$, we must have $c_2 \le 3$ since $2^43^45 = 6480 > 2013$.
* If $c_2 = 3$, we have $c_1 = 3$ since $2^43^35 = 2160 > 2013$. When $a = p_1^3 p_2^3 p_3$, we have $d(a^2) = 7^2 \cdot 3 = 147$.
* If $c_2 = 2$, we have $c_1 \le 5$ since $2^63^25 = 2880 > 2013$. When $a = p_1^5 p_2^2 p_3$, we have $d(a^2) = 11 \cdot 5 \cdot 3 = 165$.
* If $c_2 = 1$, we have $c_1 \le 7$ since $2^83 \cdot 5 = 3840 > 2013$. When $a = p_1^7 p_2 p_3$, we have $d(a^2) = 15 \cdot 3^2 = 135$.
• If $s = 2$, we must have $c_2 \le 4$ since $2^53^5 = 7776 > 2013$.
– If $c_2 = 4$, we have $c_1 = 4$ since $2^53^4 = 2592 > 2013$. When $a = p_1^4 p_2^4$, we have $d(a^2) = 9^2 = 81$.
– If $c_2 = 3$, we have $c_1 \le 6$ since $2^73^3 = 3456 > 2013$. When $a = p_1^6 p_2^3$, we have $d(a^2) = 13 \cdot 7 = 91$.
– If $c_2 = 2$, we have $c_1 \le 7$ since $2^83^2 = 2304 > 2013$. When $a = p_1^7 p_2^2$, we have $d(a^2) = 15 \cdot 5 = 75$.
– If $c_2 = 1$, we have $c_1 \le 9$ since $2^{10}3 = 3072 > 2013$. When $a = p_1^9 p_2$, we have $d(a^2) = 19 \cdot 3 = 57$.
• If $s \le 1$, we have $c_1 \le 10$ since $2^{11} = 2048 > 2013$. When $a = p_1^{10}$, we have $d(a^2) = 21$.
Therefore, $a^2$ has at most 243 positive divisors. This holds when $a = p_1^4 p_2 p_3 p_4$ for some distinct primes $p_1, p_2, p_3, p_4$. As $3^42 \cdot 3 \cdot 5 = 2430 > 2013$, we must have $p_1 = 2$. Also, since $2^43 \cdot 5 \cdot 11 = 2640 > 2013$, the only possibility is $a = 2^43 \cdot 5 \cdot 7 = 1680$.