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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 and be polynomials with complex coefficients and let be a complex number. Prove that if
for all , then the polynomials and are constants.
Solution: First, note that if is a constant, then this forces to be a constant, and vice-versa. Now, suppose and are not constants. We show that this leads to a contradiction.
Observe that if and have a common root, say, , then we have , which implies , which implies , a contradiction. Therefore, we conclude and are relatively prime polynomials, and hence, and are also relatively prime. Now, let and . Then, from the given equation, we conclude and for some .
Now, applying the Mason-Stothers theorem, we get
, which implies , 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 and are polynomials, then
- denotes the degree of ,
- denotes the number of distinct zeros of , and
- denotes the .
To illustrate, suppose and . Then, and . And, and . Also, .
Let us first prove a couple of useful lemmas before stating the theorem and its proof.
Lemma 1: If is a polynomial, then has repeated roots iff and have a common root.
Proof: If has a root of multiplicity , then , where . Therefore,
Now, if , then , which is the same thing as saying, if does not have a repeated root, then and don’t have a common root. And, if , then has root of multiplicity at least . This is same as saying, if has a repeated root , then and have a common root . And, this completes our proof.
Lemma 2: If is a polynomial, then .
Proof: Suppose and has distinct roots , with multiplicities , respectively. Then, . Now, from the proof of lemma above, we note that . Therefore,
. And, we are done.
Okay, we are now ready to state the theorem.
Mason-Stothers Theorem: If are relatively prime polynomials, not all constant, such that , then .
Proof: We first note that
Indeed, we have . Therefore,
. And, we are done.
Also, note that at least two of the polynomials and 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 and are non-constant polynomials. Now, we note that , for otherwise, , and since and are relatively prime, this would imply , which leads to a contradiction!
Now, we observe that and divide the left hand side of , and divides the right hand side of , which is equal to the left hand side. And, since and are relatively prime, we conclude that
The above implies that
Now, applying lemma to , we obtain,
Using the above in , we get
since are relatively prime polynomials.
Due to symmetricity, the above arguments can similarly be repeated for and to get similar inequalities for and . 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.)