The Universe of Discourse

Irreducible polynomials

A puzzle for high school math students

Solution

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?

[ Addendum 20220221: I accidentally wrote the same article a second time. ]

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