Multiplying numbers in the Middle Ages was considered difficult. The solution to this may surprise you.
Number theory is a near and dear subject to me, primarily due to it's purity of formulation. At the end of the day, you can clearly and easily test your results (in most cases), and once something is proven it stays proven indefinitely (note things like Fermat's Last Theorem and Riemann's Hypothesis were never proven and while people claimed to prove them they weren't considered proofs).
That said, enter Nicomachus. This guy attempted to sum up all of greek mathematics into a single volume, and he actually did a pretty good job. However, some confusing assertions come to light. Take the Regula Nicomachi as a prime example. It is noted that during the middle ages multiplying two numbers was a very difficult process. I guess they didn't have the same methods we teach our 2nd graders way back then. So they had to come up with little processes to help them. And here is one such process.
If three terms in progression are a - k, a, a + k then (a-k)(a+k) + (k*k) = (a*a), note I can't represent squaring very easily here.
The example given is for 98 and proceeds what is the value of 98 squared. Okay, plug in using k = 2, and you get.
(98 - 2)(98 + 2) + (2*2) = (98*98)
Now, I'll be damned, but that new equation has TWO multiplications and the old equation only had a single equation. Seems to me that the middle ages had some difficult times with numbers because they did too MUCH work. But wait and let's see how this pans out:
96*100 + 4 = 98 squared, wow multiplying by 100 is easier because of the 0's
98 squared = 9604
Hopefully you start to see how they used this process. The simplified the multiplication process by simplifying the numbers involved. There may be more of them, but they are now easier to work with. I mean any squire back then should have been able to multiply 96 * 100 and get 9600... I was trying to think of how to generalize this to any situation. Basically, given any number that you need to square, for which ones does the math involved become simpler using the progression and for which does it become more difficult. Take something really large like, like 3456 and square that behemoth. For us, this would take 4 single digit mults, and 1 large addition. I'm not sure which processes they had back then and which they didn't, so to them, the process was probably far more complex. In our case the numbers we need to simplify the equation live are k = 456, that is still huge, so maybe we could do something simpler and expand the problem out?
(3456*3456) = (3456 - 456)(3456+456) + (456*456), (3000)(3912) + (456*456)
(456*456) = (456 - 44)(456+44) + (44*44), (412)(500) + (44*44)
(44*44) = (44-4)(44+4) + (4*4) = 1936
(456*456) = 206000 + 1936 = 207936
(3456*3456) = 11736000 + 207936 = 11943936
Note, we are reducing the relative complexity in each multiplication to a single term at each step, and then solving for the complex k squared term by using another progression. By doing back substitution, we can eventually find the complex answer. May seem complicated to us because of our new tools for solving these types of problems, but you better bet that they found these equations much easier to use.