So I've been doing what I imagine most mathematicians have been doing during the lockdown, prowling through old journals (1770 - 1970 mostly) a few old blogs and web pages I hadn't scanned in a while, and catching up when I could on who's tweeting what in math and history... and it seems that the things leaping out at me are all about polynomials, at least mostly.
So reading through my notes about Leonard of Pisa (I hate to call him Fibonacci, no body did when he was alive, except maybe himself once but I'm feeling doubtful. And if you want to dive down that rabbit hole, here are my notes to start you off with "Who Was Fibonacci before He Was Fibonacci." )
Leonardo's most famous book was the 1202 Liber Abaci, first because it had the rabbit problem, and second because it was really important in introducing Arabic base ten numbers to the West. Perhaps number two on his hit parade was his 1225 Libre Quadratorum, Book of Squares.
Some of what he writes was known to the Pythagoreans, Proclus, and Eulcid, and almost all was known by the time of Diaphontus of Alexandria (about 100 AD). Certianly the proposition VI that I focus on was known to him, and to Brahamagupta, and in 1225 to Leonardo. Then it shows up in my other reading, but first, Proposition VI. It is sometimes called the Brahmagupta-Fibonacci identity, because Brahamagupta extend it even farther than this method, but the basic version is this.
If you have four numbers, and in the Pisano's words, "not in proportion, The first less than the second, and the third less than the fourth.." then, in modern terminology, The sum of the squares of the first two, times the sum of the squares of the last two, then the product will be expressible in two different ways as the sum of two squares. And he tells you how to get those two numbers. If the first numbers are a, b, c, and d, then (a^2 + b^2) (c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2, and don't worry about the signs, because it works if you switch them (ac-bd)^2 + (ad+bc)^2.
Now it doesn't matter much what you start with , so let's do 1, 2, 4, 7 'cause they are easy. The product of the sums of squares is (1^1 + 2^2)(4^2 +7^2) = 5(65) = 325. So the first way to get this as a sum of two squares is (1*4+2*7)^2 + (1*7-4*2)^2= 18^2 +(-1)^2 = (Eureka) 325.
Ok, but it has to work the other way, so (1*4-2*7)^2 + (1*7+4*2)^2 = (-10)^2 + 15^2 = 100 + 225 = 325....again.
Immediately you recognize (of course you would, even I did) that if you did this with two numbers that were legs of Pythagorean right triangles, you could find two right triangles that had a common hypotenuse that was the product of the two original hypotenii. And someone else who did that was the brilliant Francois Viete, around 1570 in his Genesis triagulorum.
Now I don't thing Diophantus or Fibonacci either worked with negative numbers, and neither did Viete, but he did throw a curve ball because well after Robert Record gave that perfectly good, and really long, equal sign in his Whetstone of Witte, 1557, Viete and some continental renegades continued to use the same symbol for the absolute difference of two numbers, so when he wrote a=b he meant what's the difference between the bigger and the smaller. But he wrote the same rules.
Now I admit when I first used this my mind went right to FOIL (I was lucky to be teaching before the mnemonic was sin), and the ac +bd chimed "first and last" to me, and ad + bc was outside inside, and that's how I remembered it.
And then reading Viete (translated thank you) it looked different, and for some reason for the first time I saw what everyone may be seeing who is clever, but I never had.
But before you actually think it, remember who we are reading here, Diaphontus, Leonardo de Pisa, Francois Viete, all of them living well before it was understood how to multiply and divide COMPLEX NUMBERS.
That's right, if like me it didn't hit you, let's rewrite our first problem as (1+2i)(4+7i) Multiply away using the proper distributive property without an unsavory mnemonic, 1*4 + 2i*4 + 1*7i + 2i*7i, and we get 4 + 8i + 7i - 14, or -10 + 15i, and the (modulus, magnitude, absolute value, take your choice is \(\sqrt{(-10)^2 + 15^2} |) And the other solution is just to mulitply by the conjugate of the second (1+2i)(4-7i) = (18 + 1i)
Now Viete noticed something else "complex" about those numbers. And he described the two techniques with different names, the second (First and Last negative in my example) he called synaereseos, from a Greek verb meaning 'taken together'. The first (First and Last positive), he referred to as diaeresos, meaning 'taken to pieces.' He stated clearly that if you took the difference between the two smaller angles in the first two triangles, In this case, the arctan(1/2) = 26.56 degrees; and arctan(4/7) = 29.74 degrees. Not much difference, but then by his diaeresos, or conjugate method, the result is arctan(1/18) = 3.1798 degrees.
Because he focused on the smallest angle, and didn't care whether it was next to the x or the y axis, he didn't notice that had he stayed with the angle next to the x-axis for his reference, in the Synaereseos result, the angle would have been the sum 56.31 degrees of the two original angles.
But then he broke Leonardo of Pisa's first rule. Remember he said pick four numbers not in proportion. Veite said what if we make the two sets equal, that is, square the number. Using (3,4) and (3,4), using his first (taking together) we get (9+16)^2 +(12-12)^2 =25^2 which sort of gets the answer, but the (taken apart) method gives (9-16)^2 + (12+12)^2 = 7^2 + 24^ 2 = 25^2. If we preserve the angle of reference to be the one between the x axis and the hypotenuse, then the original 3,4,5 triangles have a measure of 36.869 degrees, their sum should be 73.739 degrees,exactly arctan(24/7) and their difference should be zero, perfect for the x=7, y=0.
You get the same answers if you do these by (3+4i)^2; and it works in general for (3+4i)^n of your choice.
Because I was looking at some old blogs of Prof Dan Kalman and he had some really clever notes about.... polynomials. Some of which reminded me of some notes I wrote after reading one of his blogs back in 2013. Some of it I will share here, ignoring all the parts that required some small amount of calculus; but if you want the whole enchilada, about what Professor Kalman calls the "Marden's Theorem":
I once read a description of math as like seeing islands in a great ocean covered by a mist. As you learn the subject you work around on an island and clear away some of the mist. Often your education jumps from one island to another at the direction of a teacher and eventually you have mental maps of parts of many separate islands. But at some point, you clear away a fog on part of an island and see it connects off to another island you had partially explored, and now you know something deeper about both islands and the connectedness of math.
I was recently reminded of one of those kinds of connections that ties together several varied topics from the high school education of most good math students. It starts with that over-criticized (and under-appreciated, Algebra I technique of factoring.
Almost ever student in introductory algebra is introduced to a "sum and product" rule that relates the factors of a simple quadratic (with quadratic coefficient of one) to the coefficients. The rule says that if the roots are at p and q, then the linear coefficient will be the negative of p+q, and the constant term will be their product, pq. So for example, the simple quadratic with roots at x=2 and x=3 will be x2 - 5x + 6.
I know from experience that if you take a cross section of 100 students who enter calculus classes after two+ year of algebra, very few will know that you can extend that idea out to cubics and higher power polynomials. An example for a polynomial with four roots will probably suffice for most to understand. Because the constant terms in linear factors are always the opposite of the roots, {if 3 is a root, (x-3) is a factor} it is easiest to negate all the roots before doing the math involved (at least for me it always was). (addendum: If you're starting without knowing the roots, the rule of thumb is to change the sign of the coefficients that modify an odd power, like the x, x^3 etc.)
So if we wanted to find the simple polynomial with roots at -1, -2, -3, and -4 (chosen so all the multipliers are +, the factors are (x+1)(x+2)(x+3)(x+4) we would find that the fourth degree polynomial will have 10 for the coefficient of x3 because 1+2+3+4 = 10, just as it works in the second term of a quadratic. After that, the method starts to combine sets of them. The next coefficient will be the sum of the products of each pair of factor coefficients. In the example I created we would add 1x2+1x3+1x4+2x3+2x4+3x4 to get 35x2. The next term sums all triple products of the numbers, 1x2x3 + 1x2x4 + 1x3x4 + 2x3x4 = 50 for the linear coefficient. And in the constant term, we simply multiply all of them together to get 24.
In a talk in 2011 (I think) Prof. Kelman introduced what I guess I would call a rule of thumb about "reverse polynomials". Most of us are better at polynomials that have a first term of one, but sometimes you get some with "ugly" leading coefficients and 1 as the constant, Maybe 12x^2 - 7x + 1, and we need to find the roots. One of the nice rules Prof Kalman points out is a relation between polynomials like that, and their reversal, 1x^2-7x +12, which you suddenly, almost instantly recognize as (x-4)(x-3) with roots of x=3, 4. So how does that help, well let's walk through the other one. Reversing our thinking, then it must be (4x-1)(3x-1) and so the roots are...aha, 1/4 and 1/3, the inverse of the reversal. Sure enough, the sum of the roots is 7/12 and the product is 1/12. The professor points out one more way you can get clues about finding, or checking solutions. If you look at some big long polynomial like 2x^5 + 5x^4 + 3x^3 - 7x^2 + 3x +4, you not only know that the sum of the roots are -5/2, and the product of the roots is 4/2 = 2, you also can imagine the reverse and say the sum of the reciprocals is -4/3, and the product of the reciprocals is 1/2.
He extends this idea that the reverse polynomials have roots that are the inverse of the original to something he calls palindrom Polynomials, and shows how their roots must come in pairs of inverses. The entire set of notes in PDF are here.
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