, 2024

Solution:

Using the formula $1+2+\cdots+n=\frac{n(n+1)}{2}$, we get

$$\begin{aligned} \frac{a b(a b+1)}{2}-\frac{a(a+1)}{2} \cdot \frac{b(b+1)}{2} & =\frac{a b(2(a b+1)-(a+1)(b+1))}{4} \\ & =\frac{a b(a b-a-b+1)}{4} \\ & =\frac{a b(a-1)(b-1)}{4} \\ & =\frac{a(a-1)}{2} \cdot \frac{b(b-1)}{2} \end{aligned}$$

This means we can write the desired equation as

$$a(a-1) \cdot b(b-1)=4800$$

Assume $b \leq a$, so we know $b(b-1) \leq a(a-1)$, so $b(b-1)<70$. Thus, $b \leq 8$.
If $b=7$ or $b=8$, then $b(b-1)$ has a factor of 7, which 4800 does not, so $b \leq 6$.
If $b=6$ then $b(b-1)=30$, so $a(a-1)=160$, which can be seen to have no solutions.
If $b=5$ then $b(b-1)=20$, so $a(a-1)=240$, which has the solution $a=16$, giving $5+16=21$.
We need not continue since we are guaranteed only one solution, but we check the remaining cases for completeness. If $b=4$ then $a(a-1)=\frac{4800}{12}=400$, which has no solutions. If $b=3$ then $a(a-1)=\frac{4800}{6}=800$ which has no solutions. Finally, if $b=2$ then $a(a-1)=\frac{4800}{2}=2400$, which has no solutions.