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:

- 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). - 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).

**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 x

_{0}in

**C**. The function h: y → p(x

_{0}, y) is constant if and only if x

_{0}is in V.

By symmetry, for y

_{0}in

**C**, x → p(x,y

_{0}) is constant if and only if f(y

_{0}) = 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(x

_{0},y)) = f(x

_{0})f(y) is constant if and only if f(x

_{0}) = 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) = p

_{n}(x)y

^{n}+ … + p

_{1}(x)y

^{1}+ p

_{0}(x). By assumption, p

_{1},…,p

_{n}have infinitely many roots, and so must be zero. Thus p(x,y) = p

_{0}(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 = {x

_{1},…,x

_{n}}. If x is in E, then f(p(x,x

_{i})) = f(x)f(x

_{i}) = f(x

_{i}) ≠ 1, so p(x,x

_{i}) must be in

**C**\E. Therefore, for any x in

**C**, p(x,x

_{i}) can only take the values {x

_{1},…,x

_{n},p(x

_{1},x

_{i}),…,p(x

_{n},x

_{i})}, which means that x → p(x,x

_{i}) is constant. By lemma 1 x

_{i}must be in V, which is a contradiction since V is empty.

**Lemma 6**. E is non-empty.

**Proof**: Choose x

_{0}such that f(x

_{0}) ≠ 0. By lemma 1, the function y -> p(x

_{0},y) is not constant, so we can find a y

_{0}in

**C**, such that p(x

_{0},y

_{0}) = x

_{0}. Now f(x

_{0})f(y

_{0}) = f(p(x

_{0},y

_{0})) = f(x

_{0}), so f(y

_{0}) = 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:

_{n}(y)x

^{n}+ … + p

_{1}(y)x

^{1}+ p

_{0}(y).

Take y

_{0}in

**C**. If the polynomial x → p(x,y

_{0})-x is non-constant, we can find a root x

_{0}so that f(x

_{0})f(y

_{0}) = f(p(x

_{0},y

_{0})) = f(x

_{0}), implying f(y

_{0}) = 1, i.e. y

_{0}is in E. Thus for y

_{0}in

**C**\E, the polynomial x → p(x,y

_{0})-x is constant. From this we see that p

_{n}(y

_{0}) = … = p

_{2}(y

_{0}) = 0, p

_{1}(y

_{0}) = 1, and since this applies to all y

_{0}in

**C**\E, which is infinite by lemma 5, the polynomials p

_{1},…,p

_{n}must be constant with p

_{1}= 1, p

_{2}= … = p

_{n}= 0. In other words, p(x,y) = x + p

_{0}(y). By symmetry we also have p(x,y) = y + q

_{0}(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 = {x

_{1},…,x

_{n}}. p(x,y) is constant on each of the lines x = x

_{i}, and also on the lines y = x

_{i}. The lines together form a connected grid on which p is constant, so the constant along all the lines must be the same, p(x

_{i},y) = p(x,x

_{i}) = c for all x

_{i}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 x

_{i}in V and let y

_{0}be a root of the polynomial y → p(x',y)-x

_{i}. Then x

_{i}= p(x',y

_{0}) and 0 = f(x

_{i}) = f(p(x',y

_{0})), which by lemma 2 implies that y

_{0}is in V. Thus x

_{i}= p(x',y

_{0}) = 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) = q

_{n}(x)y

^{n}+ ... + q

_{1}(x)y + q

_{0}(x), where q

_{n}is not identically zero for an n>0, we see that the polynomial y → q(x,y) can only be constant if q

_{n}(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 f

_{1}(x) = f(x+c), such that f

_{1}(x)f

_{1}(y) = f(x+c)f(y+c) = f(p(x+c,y+c)) = f(k x

^{r}y

^{s}+ c) = f

_{1}(k x

^{r}y

^{s}), and substitute this function for f. We can also assume k = 1, by substituting the function f

_{2}(x) = f(x/k

^{t}), t = 1/(r+s-1), for f: f

_{2}(x)f

_{2}(y) = f(x/k

^{t})f(y/k

^{t}) = f(k

^{1-rt-st}x

^{r}y

^{s}) = f(k

^{-t}x

^{r}y

^{s}) = f

_{2}(x

^{r}y

^{s}).

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/(r

^{2}-s

^{2}) and s' = s/(r

^{2}-s

^{2}). Notice that r'r - s's = 1 and r's - rs' = 0. Set x = a

^{r'}b

^{-s'}, and y = a

^{-s'}b

^{r'}. We have:

^{r}y

^{s}) = f(a

^{r'r - s's}b

^{r's - rs'}) = f(a) = 1

^{r}x

^{s}) = f(a

^{r's - rs'}b

^{r'r - s's}) = f(b) = z

This contradiction proves that r = s. Our equation now has the form f(x)f(y) = f(x

^{r}y

^{r}). 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}1

^{r}) = f(x

^{r}y

^{r}) = f(x)f(y). This finishes the proof.