Published Friday 14. November 2014
I recently looked through my old math notes, and found some of my personal research projects. It's quite fun for me to read; in those days I had much more time to pursue my interest in math, and I actually managed to prove some interesting things. Here is an example:

Theorem. Let f be a non-constant complex function and let p be a complex polynomial in two variables. Assume f and p satisfies the following equation for all x and y in C: f(x)f(y) = f(p(x,y)), then one of the following holds:
1. f has no zeroes, and p(x,y) = x+y+c for a constant c.
Furthermore, the function g(x) = f(x-c) satisfies g(x)g(y) = g(x + y).
2. f has exactly one zero, and p(x,y) = k (x-c)r (y-c)s + c for complex constants k, c, and positive integers r, s.
If f is not the function x → [x≠c] (Iverson bracket), then r = s.
Furthermore, we can find constants a and b such that the function g(x) = f(ax+b) satisfies g(x)g(y) = g(xy).
The proof of the theorem uses the fact that a complex polynomial is either constant or takes all values in C. This follows directly from the Fundamental Theorem of Algebra, which says that every non-constant complex polynomial has a root.

Define V = {x ∈ C: f(x) = 0} and E = {x ∈ C: f(x) = 1}. Since f is not constant neither C\V nor C\E can be empty.

Lemma 1. Fix x0 in C. The function h: y → p(x0, y) is constant if and only if x0 is in V.
By symmetry, for y0 in C, x → p(x,y0) is constant if and only if f(y0) = 0.

Proof: Since h is a polynomial, h is either constant or takes all values in C. Since f is not constant, h is constant if and only if y → f(h(y)) is constant. The function y → f(h(y)) = f(p(x0,y)) = f(x0)f(y) is constant if and only if f(x0) = 0.

Lemma 2. p(x,y) is in V if and only if x is in V or y is in V.

Proof: p(x,y) is in V if and only if f(p(x,y)) = 0, which happens if and only if f(x)f(y) = 0, which again happens if and only if x is in V or y is in V.

Lemma 3. If y → p(x,y) is constant for infinitely many x, then p does not depend on y.

Proof: Write p as a polynomial in y with coefficients that are polynomials in x: p(x,y) = pn(x)yn + … + p1(x)y1 + p0(x). By assumption, p1,…,pn have infinitely many roots, and so must be zero. Thus p(x,y) = p0(x), which does not depend on y.

Lemma 4. V is finite.

Proof: If V is infinite, then by lemma 1 and lemma 3 we have that p(x,y) does not depend on y. Taking x' in C\V, we have f(y) = f(p(x',y))/f(x'), which does not depend on y, making f constant, which is a contradiction.

Lemma 5. If f has no zeroes, then C\E is infinite.

Proof: Assume C\E is finite, and let C\E = {x1,…,xn}. If x is in E, then f(p(x,xi)) = f(x)f(xi) = f(xi) ≠ 1, so p(x,xi) must be in C\E. Therefore, for any x in C, p(x,xi) can only take the values {x1,…,xn,p(x1,xi),…,p(xn,xi)}, which means that x → p(x,xi) is constant. By lemma 1 xi must be in V, which is a contradiction since V is empty.

Lemma 6. E is non-empty.

Proof: Choose x0 such that f(x0) ≠ 0. By lemma 1, the function y -> p(x0,y) is not constant, so we can find a y0 in C, such that p(x0,y0) = x0. Now f(x0)f(y0) = f(p(x0,y0)) = f(x0), so f(y0) = 1.

Proof of the theorem: For the first part of the theorem, assume that f has no zeroes. Write p as a polynomial in x with coefficients that are polynomials in y:

p(x,y) = pn(y)xn + … + p1(y)x1 + p0(y).

Take y0 in C. If the polynomial x → p(x,y0)-x is non-constant, we can find a root x0 so that f(x0)f(y0) = f(p(x0,y0)) = f(x0), implying f(y0) = 1, i.e. y0 is in E. Thus for y0 in C\E, the polynomial x → p(x,y0)-x is constant. From this we see that pn(y0) = … = p2(y0) = 0, p1(y0) = 1, and since this applies to all y0 in C\E, which is infinite by lemma 5, the polynomials p1,…,pn must be constant with p1 = 1, p2 = … = pn = 0. In other words, p(x,y) = x + p0(y). By symmetry we also have p(x,y) = y + q0(x). Together these two facts mean that p(x,y) = x+y+c for a constant c. If we define g(x) = f(x-c), then g(x)g(y) = f(x-c)f(y-c) = f(p(x-c,y-c)) = f(x-c+y-c+c) = f(x+y-c) = g(x+y).

For the second part of the theorem, we assume that f has at least one zero. By lemma 4, V is finite, so we can write V = {x1,…,xn}. p(x,y) is constant on each of the lines x = xi, and also on the lines y = xi. The lines together form a connected grid on which p is constant, so the constant along all the lines must be the same, p(xi,y) = p(x,xi) = c for all xi in V and all x,y in C.

Take x' in C\V. By lemma 1, the polynomial y → p(x',y) is not constant. Take any xi in V and let y0 be a root of the polynomial y → p(x',y)-xi. Then xi = p(x',y0) and 0 = f(xi) = f(p(x',y0)), which by lemma 2 implies that y0 is in V. Thus xi = p(x',y0) = c. This proves that c is the only element of V.

Since p(x,y)-c is zero on the lines x = c and y = c, we can conclude that p(x,y)-c = (x-c)r (y-c)s q(x,y) for a polynomial q, where the positive multiplicities r and s are chosen such that x-c and y-c does not divide q.

We want to prove that q is constant, so let's assume that q is non-constant in y and find a contradiction. If we write q(x,y) = qn(x)yn + ... + q1(x)y + q0(x), where qn is not identically zero for an n>0, we see that the polynomial y → q(x,y) can only be constant if qn(x) = 0, which only happens for finitely many x. Using polynomial division, we can also write q(x,y) = u(x,y)(y-c) + r(x). If q(x,y) = 0, we must have x=c or y=c, so if we take an x ≠ c such that y → q(x,y) is not constant, then c must be a root, and 0 = q(x,c) = r(x). Every such x must be a root in r(x), so r must be identically zero, and so y-c divides q. This contradicts the assumption, so q must be constant in y. The same argument proves that q is constant in x, so we have p(x,y) = k(x-c)r (y-c)s + c for a constant k.

We now prove that we can choose g(x) = f(ax+b) such that g(x)g(y) = g(xy). First of all, we can assume without loss of generality that c = 0, since we can define f1(x) = f(x+c), such that f1(x)f1(y) = f(x+c)f(y+c) = f(p(x+c,y+c)) = f(k xr ys + c) = f1(k xr ys), and substitute this function for f. We can also assume k = 1, by substituting the function f2(x) = f(x/kt), t = 1/(r+s-1), for f: f2(x)f2(y) = f(x/kt)f(y/kt) = f(k1-rt-st xr ys) = f(k-t xr ys) = f2(xr ys).

By lemma 6 we know that 1 is contained in f(C), so f(C) will at least contain {0,1}. If f(C) = {0,1}, then there is only one possibility for f, since we must have f(x)=0 if and only if x=0. This function can be written x → [x≠0]. It is obvious that f(x)f(y) = f(xy).

Now we can assume that f(C) contains an element z not in {0,1}. First we prove that r = s: Assume r > s, and choose a with f(a)=1 and b with f(b)=z:

Set r' = r/(r2-s2) and s' = s/(r2-s2). Notice that r'r - s's = 1 and r's - rs' = 0. Set x = ar' b-s', and y = a-s' br'. We have:

f(x)f(y) = f(xr ys) = f(ar'r - s's br's - rs') = f(a) = 1

f(y)f(x) = f(yr xs) = f(ar's - rs' br'r - s's) = f(b) = z

This contradiction proves that r = s. Our equation now has the form f(x)f(y) = f(xr yr). Put x=y=1 to get f(1)f(1) = f(1), and therefore f(1) = 1.

Finally we have: f(xy) = f(xy)f(1) = f((xy)r 1r) = f(xr yr) = f(x)f(y). This finishes the proof.