SFFT won't work here

First note that $3 | (2a+b)+(2b+a) = 3(a+b)$ . Given that $9 | 4752$ , we must thus have both $3 | (2a+b)$ and $3 | (2b+a)$ .

Note that $4752 = 9 \cdot 528 = 9 \cdot (23^2-1)=3 \cdot 22 \cdot 3 \cdot 24$ .

Assuming w.l.o.g. that $a<b \Rightarrow 2a+b < 2b+a$ (obviously $a \neq b$ since $4752$ is not a perfect square).

We consider when $2a+b < 3 \cdot 22$ , with the largest factor of $528$ smaller than $22$ being $16$ and concurrently the smallest factor larger than $24$ is $33$ .

Note for all such $2a+b < 3 \cdot 22$ , we have $(2a+b)+(2b+a) = 3(a+b) \geq 3\cdot(16+33) \Rightarrow a+b \geq 49$ . Yet we also have $2a+b \leq 48$ , a contradiction.

Hence we must have $2a+b = 3\cdot 22, 2b+a = 3\cdot24 \Rightarrow a=20, b=26 \Rightarrow ab = \boxed{520}$

First, take the equation $\mod 2$ to see that $ab \equiv 0 \mod 2$ so at least one of the variables is even. Without loss of generality let's assume $b = 2b_0$ and rewrite the equation as $(2a + 2b_0)(4b_0 + a) = 4752 \implies (a + b_0)(4b_0 + a) = 2376$ . Now take the equation $\mod 3$ to see that $(a + b_0)^2 \equiv 0 \mod 3 \implies a + b_o \equiv 0 \mod 3$ . Thus we may say that $a + b_0 = 3k$ . Let's rewrite the equation in terms of only $k$ and $b_0$ : $3k(3k + 3b_0) = 2376 \implies 9k(k + b_0) = 2376 \implies k(k+b_0) = 264 \implies b_0 = \frac{264}{k} - k$ where it is clear that $k | 264$ . Let's now write the original variables in terms of k:

$b = 2b_0 = \frac{528}{k} - 2k$ and $a = 3k - b_0 = 4k - \frac{264}{k}$ . It is clear that any valid choice of $k$ will give a solution to the equation, but we must first guarantee that both variables will be positive: $b > 0 \iff \frac{528}{k} > 2k \iff 264 > k^2 \iff 16 \geq k$ and $a > 0 \iff 4k > \frac{264}{k} \iff k^2 > 66 \iff k \geq 9$ so $16 \geq k \geq 9$ . But we must remember that $k | 264 = 2^3 \times 3 \times 11$ . Thus, the only valid values of $k$ are $k = 11,12$ . Choosing $k = 11$ we get $b = 26, a = 20$ . If we instead choose $k = 12$ , the order of the values is reversed so in both cases $ab = 26 \times 20 = 520$ .