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Yesterday, I wrote a post on the Mason-Stothers theorem and presented an elementary proof of the theorem given by Noah Snyder. As mentioned in that post, I will present now a problem proposed by Magkos Athanasios (Kozani, Greece) that can be solved almost “effortlessly” using the aforesaid theorem.

Problem: Let $f$ and $g$ be polynomials with complex coefficients and let $a \ne 0$ be a complex number. Prove that if

$(f(x))^3 = (g(x))^2 + a$

for all $x \in \mathbb{C}$, then the polynomials $f$ and $g$ are constants.

(Magkos Athanasios)

Solution: First, note that if $f$ is a constant, then this forces $g$ to be a constant, and vice-versa. Now, suppose $f$ and $g$ are not constants. We show that this leads to a contradiction.

Observe that if $f$ and $g$ have a common root, say, $\alpha$, then we have $(f(\alpha))^3 = (g(\alpha))^2 + a$, which implies $0 = 0 + a$, which implies $a = 0$, a contradiction. Therefore, we conclude $f, g$ and $a$ are relatively prime polynomials, and hence, $f^3, g^2$ and $a$ are also relatively prime. Now, let $\deg (f) = n$ and $\deg (g) = m$. Then, from the given equation, we conclude $n = 2k$ and $m = 3k$ for some $k \in \mathbb{N}$.

So,

$\max \{\deg(f^3), \deg(-g^2), \deg(a)\} = \max \{6k, 6k, 0\} = 6k$.

Also,

$N_0 (f^3 (-g^2) a) - 1$

$= N_0 (fg) - 1 \le \deg (f) + \deg(g) - 1 = 2k + 3k - 1 = 5k - 1$.

Now, applying the Mason-Stothers theorem, we get

$6k \le 5k - 1$, which implies $k \le -1$, a contradiction! And, we are done.

About a couple of weeks ago, Noah Snyder liveblogged at the Secret Blogging Seminar on two topology talks given by Jacob Lurie. You may want to learn more about the contents of the talks by clicking the appropriate link. The thing is when I came to know about Noah’s liveblog, the thought that immediately sprang to my mind was the one involving his elementary proof of the well-known Mason-Stothers theorem.

In this post, I wish to present Noah Snyder’s proof of the Mason-Stothers Theorem, which is the polynomial version of the yet unproven (and well-known) ABC Conjecture in number theory. I will follow it up with a problem and its solution using the aforesaid theorem. For a detailed and wonderful exposition on the ABC conjecture, you may want to read an article, titled The ABC’s of Number Theory, written by Noam Elkies for The Harvard College Mathematics Review.

First, a brief history. Though this theorem on polynomials was proved by Stothers in 1981, it didn’t attract much attention until 1983 when it was rediscovered by Mason. Noah, in 1998 (while still in high school), gave perhaps the most elegant elementary proof.

The proof below is the version given in Serge Lang’s Undergraduate Algebra, which it seems has quite a number of typos.

Okay, now some terminology. If $f$ and $g$ are polynomials, then

• $\deg(f)$ denotes the degree of $f$,
• $N_0 (f)$ denotes the number of distinct zeros of $f$, and
• $(f, g)$ denotes the $\gcd (f, g)$.

To illustrate, suppose $f(x) = x(x-1)^2 (x+2)^3$ and $g(x) = 5x^2$. Then, $\deg(f) = 6$ and $N_0 (f) = 3$. And, $\deg(g) = 2$ and $N_0 (g) = 1$. Also, $(f, g) = x$.

Let us first prove a couple of useful lemmas before stating the theorem and its proof.

Lemma 1: If $f \in \mathbb{C}[t]$ is a polynomial, then $f$ has repeated roots iff $f$ and $f'$ have a common root.

Proof: If $f$ has a root $\alpha$ of multiplicity $k$, then $f(x) = (x - \alpha)^k Q(x)$, where $Q(\alpha) \ne 0$. Therefore,

$f'(x) = k(x - \alpha)^{k-1} Q(x) + (x - \alpha)^k Q'(x)$

$= (x - \alpha)^{k-1} (kQ(x) + (x - \alpha)Q'(x)$.

Now, if $k = 1$, then $f'(\alpha) = Q(\alpha) \ne 0$, which is the same thing as saying, if $f$ does not have a repeated root, then $f$ and $f'$ don’t have a common root. And, if $k > 1$, then $f'$ has root $\alpha$ of multiplicity at least $k-1$. This is same as saying, if $f$ has a repeated root $\alpha$, then $f$ and $f'$ have a common root $\alpha$. And, this completes our proof.

Lemma 2: If $f \in \mathbb{C}[t]$ is a polynomial, then $\deg (f, f') = \deg (f) - N_0 (f)$.

Proof: Suppose $\deg (f) = n$ and $f$ has distinct roots $\alpha_1, \alpha_2, \ldots , \alpha_i$, with multiplicities $k_1, k_2, \ldots, k_i$, respectively. Then, $n = k_1 + k_2 + \ldots + k_i$. Now, from the proof of lemma $(1)$ above, we note that $(f, f') = (x-\alpha_1)^{k_1 - 1}(x-\alpha_2)^{k_2 - 1} \ldots (x-\alpha_n)^{k_n - 1}$. Therefore,

$\deg (f, f')$

$= (k_1 - 1) + (k_2 - 1) + \ldots + (k_i - 1)$

$= (k_1 + k_2 + \ldots + k_i) - i$

$= n - i$

$= \deg (f) - N_0 (f)$. And, we are done.

Okay, we are now ready to state the theorem.

Mason-Stothers Theorem: If $f, g, h \in \mathbb{C}[t]$ are relatively prime polynomials, not all constant, such that $f + g = h$, then $\max \{\deg(f), \deg(g), \deg(h)\} \le N_0 (fgh) - 1$.

Proof: We first note that

$f'g - fg' = f'h - fh' \quad \quad (*)$.

Indeed, we have $f' + g' = h'$. Therefore,

$f'g - fg' = f'(h-f) - f(h' - f') = f'h - fh'$. And, we are done.

Also, note that at least two of the polynomials $f, g$ and $h$ are non-constants, for if any two polynomials are constants, then this forces the third to be a constant as well. So, without any loss of generality, assume $f$ and $g$ are non-constant polynomials. Now, we note that $f'g - fg' \ne 0$, for otherwise, $f'g = fg' \ne 0$, and since $f$ and $g$ are relatively prime, this would imply $g \mid g'$, which leads to a contradiction!

Now, we observe that $(f, f')$ and $(g, g')$ divide the left hand side of $(*)$, and $(h, h')$ divides the right hand side of $(*)$, which is equal to the left hand side. And, since $f, g$ and $h$ are relatively prime, we conclude that

$(f, f')(g, g')(h, h')$ divides $f'g - fg'$.

The above implies that

$\deg (f, f') + \deg (g, g') + \deg (h, h') \le \deg (f'g - fg')$,

which implies

$\deg (f, f') + \deg (g, g') + \deg (h, h') \le \deg (f) + \deg (g) - 1 \quad \quad (**)$

Now, applying lemma $(2)$ to $f, g, h$, we obtain,

$\deg (f, f') = \deg (f) - N_0 (f)$

$\deg (g, g') = \deg (g) - N_0 (g)$

$\deg (h, h') = \deg (h) - N_0 (h)$

Using the above in $(**)$, we get

$\deg (h) \le N_0 (f) + N_0 (g) + N_0 (h) - 1 = N_0 (fgh) - 1$

since $f, g, h$ are relatively prime polynomials.

Due to symmetricity, the above arguments can similarly be repeated for $f$ and $g$ to get similar inequalities for $\deg (f)$ and $\deg(g)$. And, this concludes our proof.

( Earlier, I had mentioned I would pose a problem and also give its solution that would use the above theorem. I will do that in my next post.)

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