# Forbidden polynomial identities by the abc conjecture

The Mason–Stothers theorem states

Let $a(t), b(t)$, and $c(t)$ be relatively prime polynomials such that $a + b = c$, with coefficients that are either real numbers or complex numbers. Then $\max\{\deg(a),\deg(b),\deg(c)\} \le \deg(\operatorname{rad}(abc))-1$

It is generalized to multivariate polynomials. The bound is tight for 3 polynomials in the sense that there are examples $\deg(c)=\deg(\operatorname{rad}(abc)) - 1$.

The abc conjecture for integers appears to forbid some polynomial identities for multivariate polynomials that are seemingly allowed by the polynomial abc theorems.

Let $a,b,c$ be relatively prime polynomials with integer coefficients in $x,y$ and $a+b=c$.

Suppose $\deg(c) = \deg(\operatorname{rad}(abc)) - 1$ and $x y$ divides $\operatorname{rad}(abc)$. Setting $x,y$ to large coprime integer powers in practice will make $x y$ vanish in the radical, $c = O(x^{(\operatorname{rad}(a(x,y)b(x,y)c(x,y)) - 1)})$ while $\operatorname{rad}(abc)=O(x^{(\operatorname{rad}(a(x,y)b(x,y)c(x,y)) - 2)})$ in a sense decreasing the "degree" of the radical by $2$ and giving infinite family of counterexamples to abc.

Another forbidden identity appears $\deg(c) = \deg(\operatorname{rad}(abc)) - 2$ and $x y (x^2-y^2-1)$ divides $\operatorname{rad}(abc)$ and $(x^2-y^2-1)$ doesn't divide $c$. In $x y (x^2-y^2-1)$ setting $x,y$ to consecutive powerful numbers $s+1,s$ gives upper bound for the radical $\operatorname{rad}(2 s (s+1)) \sim s = y$ which reduces the "degree" of the radical by $3$, again contradicting abc.

Another one appears $\deg(c) = \deg(\operatorname{rad}(abc)) - 1$ and $x^4 + k y^4$ divides $abc$ and the elliptic curve $u^4 + k = v^2$ is of positive rank.

I suppose the upper bound for the radical is OK in both cases, but not sure if the estimate for $c$ is correct -- some cancellation might make $c$ smaller, that is why the restriction on divisibility of $c$.

(A potential problem might be $a,b$ not being coprime in integers, but clearing the gcd might solve it, not entirely sure about this).

Question:

Does the abc conjecture really imply the above identities don't exists? (My reasoning well might be wrong).

Subquestion in case the main question is true:

Is there an unconditional proof that these polynomial identities don't exist?

EDIT: In comments Peter Mueller claimed that the slight modification of the second identity $x y (x^2-y^2)$ divides $a b$, $\deg(c)=\deg(\operatorname{rad}(abc)) - 2$ does not exist too.

Here is a counterexample to his claim (in machine readable form, the degrees are high)

a=(16384) * y^6 * x^6 * (x - y)^6 * (x + y)^6 * (x^2 + y^2)^6 * (x^4 - 10*x^2*y^2 + y^4)^3 * (x^8 - 8*x^6*y^2 + 30*x^4*y^4 - 8*x^2*y^6 + y^8)^3 * (x^8 + 4*x^6*y^2 + 54*x^4*y^4 + 4*x^2*y^6 + y^8)^3
b=(x^8 - 20*x^6*y^2 + 6*x^4*y^4 - 20*x^2*y^6 + y^8)^2 * (x^8 - 4*x^6*y^2 + 38*x^4*y^4 - 4*x^2*y^6 + y^8)^2 * (x^16 - 8*x^14*y^2 - 4*x^12*y^4 + 72*x^10*y^6 + 902*x^8*y^8 + 72*x^6*y^10 - 4*x^4*y^12 - 8*x^2*y^14 + y^16)^2 * (x^16 + 8*x^14*y^2 + 220*x^12*y^4 - 1480*x^10*y^6 + 3526*x^8*y^8 - 1480*x^6*y^10 + 220*x^4*y^12 + 8*x^2*y^14 + y^16)^2
c=a+b=(x^8 + 4*x^6*y^2 - 42*x^4*y^4 + 4*x^2*y^6 + y^8)^4 * (x^16 - 16*x^14*y^2 + 172*x^12*y^4 - 304*x^10*y^6 + 1318*x^8*y^8 - 304*x^6*y^10 + 172*x^4*y^12 - 16*x^2*y^14 + y^16)^4


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Probably I am wrong about at least the second identity and it well might be possible. – joro Nov 18 '12 at 14:17
1. It would be convenient, for background, if you would include in your question what the generalization to multivariable polynomials is and in what sense it is tight. 2. I don't think the word "identity" is correct in the way you are using it, since conditions on the degrees or factors of polynomials are not usually called identities. It would be better to speak about degree and divisibility conditions on polynomials instead of polynomial identities. – KConrad Nov 18 '12 at 18:28
@joro: when I asked about what sense the bound is tight, I meant the generalization, not just the case of 3 polynomials. My impression was that when you have more than 3 polynomials it is still an open question whether the bound in the conjecture is achieved. – KConrad Nov 19 '12 at 21:36
@KConrad I meant 3 polynomials, edited the question. – joro Nov 20 '12 at 6:44
There is a much simpler counterexample (to my claim in a comment) than your degree $96$ construction, coming from the parameterization of the Pythagorean triples: $a=(2xy)^2$, $b=(x^2-y^2)^2$, $c=(x^2+y^2)^2$. Here $deg(c)=4$, while $deg(rad(abc))=6$. My answer below isn't quite correct, I have to fix that later. – Peter Mueller Nov 22 '12 at 11:25

EDIT: I removed my wrong answer, I'm not even claiming anymore that the first case (where $xy$ divides $\text{rad}(abc)$) does not exist. Still, I believe that tricks like setting $y=\lambda x$ for a suitable complex number $\lambda$ and applying the univariate ABC inequality might help here. However, my original argument suffers if too high powers of $x$ divide $a(x,\lambda x)$ and $b(x,\lambda x)$.
Thank you. Do you think if in the second identity one drops $-1$, i.e. $x y (x^2-y^2)$ divides $ab$ it will still be forbidden? (You didn't use $-1$ in your proof). – joro Nov 21 '12 at 14:54
@joro: Yes, certainly for the factor $xy(x^2-y^2)$, because upon setting $y=\lambda x$, the $\text{rad}$ of $xy$ and $x^2-y^2$ drop. In your original question, you had the factor $xy(x^2-y^2-1)$ though. So what is the right version you are interested in? – Peter Mueller Nov 22 '12 at 6:11
The identity without $-1$ certainly exists, edited the question giving an example. – joro Nov 22 '12 at 7:06