Another thought about the problem of showing that every curve has an inscribed square. My guess is that more is true: that in fact every closed curve in the plane has an *odd number* of inscribed squares. In particular, this number cannot be zero!

Why make this guess? Because this is often the kind of behavior that one sees when trying to solve problems concerning real numbers. The real numbers are great for certain things (like measuring quantities of real objects!) but from any healthy mathematical perspective they are a bit of a mess. The complex numbers are much nicer.

Here’s an analogy that might be convincing. Start with this fact: if P(x) is a polynomial of odd degree, then P(x) has a *root:* that is, there exists some real number x such that P(x) = 0. How do you know this? Well, write

P(x) = a_n x^n + …. + a_1 x + a_0.

and suppose that a_n is positive. Then for x very, very large, the a_n x^n term will be much larger than all the others put together, and P(x) will be positive. But for x very, very small (that is, a negative number very far from 0), the a_n x^n term will be much _smaller_ than all the others put together, and P(x) will be negative. But if P(x) is negative for small x and positive for large x, it must be 0 somewhere in between!

But more is true: not only does there exist a root, *the number of roots is odd.* The argument is exactly the same. On the far left end of the number line, P(x) is negative. On the far right end, P(x) is positive. Thus, P(x) must switch signs an odd number of times as x moves from left to right. But each sign-switch is a point x where P(x) = 0, which is to say, a root. (Eagle-eyed readers will note that I’ve glossed over the issue of what happens when P is tangent to the x-axis at some point.) Anyway, I looked this up on MathSciNet to see if it was reasonable and found a 1961 paper of R.P. Jerrard which proves that, for plane curves defined by real-analytic functions (never mind the definition if you don’t know it, just take it to mean “in an interesting special case”) there are indeed an odd number of inscribed squares.

**Question:** Is there an easy proof that, say, a triangle has an odd number of inscribed squares? What about a quadrilateral? (One should be more careful and say “generic” triangle and “generic” quadrilaterals; there ought to be certain special choices of polygon for which the answer is negative. For instance, rectangles have *infinitely many* inscribed squares, so the question doesn’t even make sense.)

**Update:** WordPress seems to have eaten part of this post for reasons I don’t quite understand. Now fixed.

Hmm, this post seems to have gone slightly pear-shaped near the end.