This is a rant. Remember the title of the blog. You have been warned.
It has always slightly bothered me how sometimes simple arguments are made unnecessarily complicated. Sometimes people do this on purpose, to make themselves look smart. More often, people do this inadvertently, by assuming that, once they have understood something, their way is the only way. This is why, in maths, there is a difference between a proof and an elegant proof, and looking for simpler ways to prove known results is part of the job. Other disciplines (which shall remain unnamed but most readers can easily figure out) could learn a thing or two from that.
Anyway, this seems to be becoming a bit of a trend nowadays. Take something easy, implicitly (or explicitly!) claim it is complicated, and show a way-too-complex solution. One example bothered me a bit today. I ran across a youtube video (which I will not link to) solving the following problem, allegedly from the International Mathematical Olympiad: find all integer triples such that adding any of the three integers to the product of the other two yields exactly two. OK, I thought. First, I doubt this is an IMO problem: it is way too simple (see the actual problems here; participants are given 4.5 hours to solve six of them). Second, I scanned the video and it needed a full eight minutes including third-degree polynomials and “long divisions” of polynomials. Really? Sorry, but that is not elegant. Actually, the solution to this particular problem fits in a post-it (I know because that’s how I did it), and illustrates a classical way of thinking (as opposed to blindly computing). So, just getting it off my chest while having a morning coffee. Skip to the end if the problem is obvious for you.
Step one, formalize. The problem is to find all the integer triples (x,y,z) such that x+yz=y+xz=z+xy=2. Yawn.
Step two, think (and reduce to a simpler problem). One obvious solution is x=y=z=1. Duh. So the problem is: are there any other ones? Actually, it is even easier. If x=1, the system reduces to yz=1 and y+z=2, which implies y=z=1 (since they have to be integers). Analogously if y=1 or z=1. So we conclude that if x=1 or y=1 or z=1, then x=y=z=1.
Step three, solve the simpler problem. We are already left with only one case: Is there any solution to the system of equations such that neither x, nor y, nor z are equal to 1? Well, if you subtract y+xz=2 from x+yz=2 you get (x-y)(1-z)=0, and, since z is not 1, you have that x=y. Analogously, if you subtract z+xy=2 from x+yz=2 you get (x-z)(1-y)=0, and, since y is not 1, you have that x=z. So x=y=z. But then the entire system of equations reduces to a second-degree polynomial (not a third-degree one, youtubers!), x2+x-2=0. Now you can either use the standard high-school formula for the roots of a second-degree polynomial, or just note that x=1 is an obvious root and apply Ruffini’s rule (seriously: has everybody forgotten Ruffini’s rule?), or simply remember that a second-degree polynomial is x2-Sx+P, where S and P are the sum and the product of the two roots, respectively… so this polynomial has the roots x=1 and x=-2. But since we are in the case where x is not 1, and we know x=y=z, we conclude that the only other integer triple solving the problem is x=y=z=-2. Done.
And maybe there is an even simpler and more elegant solution. My point is that sometimes, after you have solved a problem (in maths or elsewhere), you should stop and think whether there is a simpler way. Especially before going off and making a full youtube video about it. Always remember the golden advice: Stop and Think!
Carlos' Ramblings 