Mon, 29 May 2006
A puzzle for high school math studentsFactor x4 + 1.
SolutionAs I mentioned in yesterday's very long article about GF(2n), there is only one interesting polynomial over the reals that is irreducible, namely x2 + 1. Once you make up a zero for this irreducible polynomial, name it i, and insert it into the reals, you have the complex numbers, and you are done, because the fundmental theorem of algebra says that every polynomial has a zero in the complex numbers, and therefore no polynomials are irreducible. Once you've inserted i, there is nothing else you need to insert.
This implies, however, that there are no irreducible real polynomials of degree higher than 2. Every real polynomial of degree 3 or higher factors into simpler factors.
For odd-degree polynomials, this is not surprising, since every such polynomial has a real zero. But I remember the day, surprisingly late in my life, when I realized that it also implies that x4 + 1 must be factorable. I was so used to x2 + 1 being irreducible, and I had imagined that x4 + 1 must be irreducible also. x2 + bx + 1 factors if and only if |b| ≥ 2, and I had imagined that something similar would hold for x4 + 1—But no, I was completely wrong.
x4 + 1 has four zeroes, which are the fourth roots of -1, or, if you prefer, the two square roots of each of i and -i. When I was in elementary school I used to puzzle about the square root of i; I hadn't figured out yet that it was ±(1 + i)/√2; let's call these two numbers j and -j. The square roots of -i are the conjugates of j and -j: ±(1 - i)/√2, which we can call k and -k because my typewriter doesn't have an overbar.
So x4 + 1 factors as (x + j)(x - j)(x + k)(x - k). But this doesn't count because j and k are complex. We get the real factorization by remembering that j and k are conjugate, so that (x - j)(x - k) is completely real; it equals x2 - √2x + 1. And similarly (x + j)(x + k) = x2 + √2x + 1.
So x4 + 1 factors into (x2 - √2x + 1)(x2 + √2x + 1). Isn't that nice?