A date in history -- minimum absolute value of difference of its factors.

decompose $19450508=2^2\times 7 \times 11^2 \times 5741$ . if $\frac{19450508}{a}=b$ , then we basically need to partition the decomposition into two groups. let's assume the difference $a-b= 5741-2^2\times 7 \times 11^2 =2353$ . note that $5741$ should be in either $a$ or $b$ . If any other factor, that is not equal to $1$ , goes with $5741$ in the same group, then the difference becomes larger than $5741-2^2\times 7 \times 11^2 =2353$ , since $a$ becomes larger and $b$ becomes smaller. So $2353$ is the answer, as the minimum difference.

What is needed is the largest factor smaller than the square root of 19450508, which is, by the way, not a prefect square. That limit value is $\left\lfloor \sqrt{19450508}\right\rfloor \Rightarrow 4410$ . The list of factors of $n$ are 1, 2, 4, 7, 11, 14, 22, 28, 44, 77, 121, 154, 242, 308, 484, 847, 1694, 3388, 5741, 11482, 22964, 40187, 63151, 80374, 126302, 160748, 252604, 442057, 694661, 884114, 1389322, 1768228, 2778644, 4862627, 9725254 and 19450508. The factor actually needed can be found by search downwards from 4410 until the factor is found: $\text{Do}[\text{If}[(19450508 \bmod i)=0,\text{Print}[i];\text{Break}[]],\{i,4410,1,-1\}] \Rightarrow 3388$

$\frac{19450508}{3388}-3388\Rightarrow 2353$