# Argument againts the $abcd$ conjecture with extra gcd conditions

Got an argument and numerical support againts the $abcd$ conjecture with extra gcd conditions (observe that this is different from the $abc$ and the $abcd$ conjectures).

This thesis p. 20 defines the $abcd$ conjecture with extra gcd conditions.

$a,b,c,d$ are pairwise coprime integers, $a+b+c+d=0$ and $d$ is positive with largest absolute value.

$$\lim\sup \frac{\log(d)}{\log\left({\rm rad}(abcd)\right)}= 1$$

Couldn't find reference in the thesis for this claim.

Q1 What is a reference for this conjecture?

The argument against it. From an identity for sum of two fourth powers define:

a(t)=3*t^13 - 38*t^12 + 516*t^11 - 4240*t^10 + 20208*t^9 - 63744*t^8 + 144960*t^7 - 248192*t^6 + 325632*t^5 - 328704*t^4 + 251904*t^3 - 139264*t^2 + 49152*t - 8192
b(t)=t^13 - 40*t^12 + 188*t^11 + 472*t^10 - 9456*t^9 + 50496*t^8 - 155072*t^7 + 316160*t^6 - 454656*t^5 + 473088*t^4 - 354304*t^3 + 182272*t^2 - 57344*t + 8192
c(t)=t^13 + 14*t^12 - 460*t^11 + 3664*t^10 - 17616*t^9 + 62208*t^8 - 166976*t^7 + 334208*t^6 - 488448*t^5 + 513024*t^4 - 378880*t^3 + 188416*t^2 - 57344*t + 8192
d(t)=3*t^13 - 40*t^12 + 540*t^11 - 3944*t^10 + 16368*t^9 - 40512*t^8 + 55488*t^7 - 15616*t^6 - 92160*t^5 + 196608*t^4 - 208896*t^3 + 133120*t^2 - 49152*t + 8192


They are pairwise coprime and satisfy

$$a(t)^4+b(t)^4-c(t)^4-d(t)^4=0$$

By the degree argument, for fixed $t$ whenever they are coprime the abcd quadruple is of good quality. $\deg (d(t)^4) = \deg (a(t)b(t)c(t)d(t))=52$.

The for integer $t$, the $\gcd$ of any of them is bounded by the resultant. The radical of the product of all resultants is $6$. Working modulo $6$, they are pairwise coprime when $t \equiv 5 \pmod{6}$.

To reduce the radical sufficiently, chose natural $u$ coprime to $6$, natural $n$ sufficiently large and solve $a(r) \equiv 0 \pmod{u^n}$. Then $a(r)=O(r^{13})$ and upper bound for the radical is $6 u O(r^{12})$, which for sufficiently large $n$ gives quality of $\frac{52}{51+o(1)}$.

To ensure coprimality with the chinese remainder theorem set $t \equiv 5 \pmod{6},t \equiv r \pmod{u^n}$.

This makes the quality $\frac{52}{51+o(1)}$ and the constant probably can be made explicit.

Experimentally, in very short time found $1520$ quadruples of quality at least $1.019$ ($\frac{52}{51}=1.0196\ldots$). To avoid factorization, gave upper bound for the radical. Some of them were with very large radical and large merit (not sure the merit makes sense in this case).

Q2 Is this really a counterexample to the conjecture?

• From my reading, it seems that this conjecture is due to the author of the thesis. Oct 31, 2014 at 15:22
• @EmilJeřábek I thought about this too. Though the author doesn't give reference for the $abc$ conjecture too, which definitely is not theirs.
– joro
Oct 31, 2014 at 15:36
• Well, the abc conjecture is famous so no one will mistake it, and the author does give a reference for the $n$-conjecture. But I agree that it is unclear. My impression is based more on the wording of e.g. section 5. If nothing else turns up, you might just try to contact him or de Weger. Oct 31, 2014 at 15:57
• As has been noted many time, by many people (especially Lang, Vojta, etc), in problems of this sort, one should allow the exclusion of a Zariski closed set of solutions. So no one-parameter family is a true counterexample. So one hopes to prove the desired result off of a closed set Z, which is a finite union of curves, and one then studies what happens on each curve, which is hopefully an easier problem, since of lower dimension. [BTW, does Vojta imply the stated result off of a closed set? It would be nice if you'd check such things.] Sep 12, 2015 at 22:27
• @JoeSilverman I am arguing against specific conjecture. Btw, neither abc nor abcd allow exceptional set. Which of Vojta's conjectures do you mean? Believe this is in the exceptional set of Vojta's "More general abc".
– joro
Sep 13, 2015 at 6:39

I am a little confused about what is being asked. If you simply look at $(T+1)^3-(T-1)^3=6T^2+2$, and substitute in $T=t^N$ then the degree of the equation is $3N$ and the degree of the radical is $2N+1$, so there can be no abcd-conjecture over the integer polynomials with exponent 1. If we then sub in integers for $t$, we see that the abcd-exponent over the integers must therefore be at least $3/2$.

The Brownawell-Masser paper states that if the polynomial terms have no common factors then one has an upper bound exponent of 2. I believe that it is an open question as to what is the best possible exponent for the polynomial conjecture (which should after all be easier than the integer conjecture).

• This question is about pairwise coprime a,b,c,d. Are yours pairwise coprime?. It is known (via different argument) that for abcd the exponent can't be smaller than $3$ (there are infinitely many quadruples with quality $>3$).
– joro
Sep 11, 2015 at 4:40
• For the polynomial case. There is identity $a(t)+b(t)+c(t)+d(t)=0$, $\deg(a(t))=3$, $\deg(rad(a(t)b(t)c(t)d(t)))=2$. The radical can easily be made $<t$ over integers, giving quality $>3$.
– joro
Sep 11, 2015 at 6:16

In the given example it is easy to pick solutions with gcd 2 in the integers, so the objection is mute.

The theorem of Brownawell-Masser for polynomials gives an exponent 2 for factors that appear in only one of $a,b,c,d$, and exponent 3 for factors that appear in two of $a,b,c,d$ (Factors can not appear in more, since we can assume gcd$(a,b,c,d)=1$). It is true that the exponent 3 is best possible here, simply taking any best-possible $a+b=c$ example and then considering $$a^3 + 3abc + b^3 = c^3.$$ However the exponent 2 is more mysterious.

I remembered that there is a better example for the coprime case: $$(T+1)^5-(T-1)^5 = 5(T^2+1)^2-8.$$ We can sub in $T=t^N+1$ to have degree $5N$ with $3N+1$ roots, so here the exponent is 5/3. We can make these almost coprime integers by selecting $t=7$ (the last two terms will have gcd 4)

• This question is about pairwise coprime a,b,c,d. It doesn't allow gcd $2$.
– joro
Sep 12, 2015 at 8:42