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Consider the identity $$ (x+z)^5+(y-z)^5 = (-3 x + 4 y)^2 (x + y)^3 + (x+y) f(x,y,z) $$

Where $f(x,y,z)=(-8*x^4 + 5*x^3*y + 24*x^2*y^2 - 9*x*y^3 - 15*y^4 + 5*x^3*z - 5*x^2*y*z + 5*x*y^2*z - 5*y^3*z + 10*x^2*z^2 - 10*x*y*z^2 + 10*y^2*z^2 + 10*x*z^3 - 10*y*z^3 + 5*z^4)$.

The curve $C : f(x,y,z)=0 $ is genus 1, have infinitely many rational and integral points since it is projective.

For a point $(x,y,z)$ on $C$, the identity becomes.

$$ (x+z)^5+(y-z)^5 = (-3 x + 4 y)^2 (x + y)^3 \qquad (1) $$

(1) has infinitely many integer solutions.

Sum of two coprime fifth powers being squarefull infinitely often contradicts the $abc$ conjecture, so $abc$ implies $\gcd(x+z,y-z) > 1$ (actually it implies the $\gcd$ is sufficiently large, since clearing a small gcd will produce abc triples of sufficiently good quality).

$C$ is birationally equivalent to $ E: y^2 = x^3 - \frac{57648010}{243}x - \frac{346032180025}{19683}$ and computing solutions to (1) gives large gcd, as implied by $abc$. Computing the gcd of the symbolic maps from the Weiersstras model give $\gcd=7$ (modulo errors).

Other similar identities exist, including genus 0 curves. The parametrization of genus 0 and $abc$ for polynomials implies common factor, though the genus 1 case is not clear to me.

Why the $\gcd$ is sufficiently large?

Is there an unconditional proof that for all similar identies the $\gcd$ will be sufficiently large? ($abc$ implies this).

Added Charles asked in a comment about points of infinite order on $C$.

Here are some:

 (27, 1, 15)
 (-1343, -1184, 279)
 (-113217, -61531, 74507)
 (-1038297, 1267624, 243888)
 (18490353467, 11046438881, 1513527591)
 (17139398481243, 15697885061884, 4151488981525)
 (-26723833000980177, 15287849768762549, 47286394561187571)
 (-1316887777770612905003, -1407701177079680302604, 837630236024655513348)

If $(-u,v)$ note the minus is on $E$ a map to (x,y,1) on $C$ is:

x,y=(-170318769169205125-1417766490*u^3-531441*u^4+1975193766630*u^2-40040866545750*v+99059869755*v*u+40507614*v*u^2)/(-601680754627470*u-63038098935*u^2+531441*u^4+1748578671*u^3+135424192071124075), (-2125764*u^4-4017005055*u^3+4664819321190*u^2-261600328098900*u-279987440048888425-91293175724310*v+243146953035*v*u+60761421*v*u^2)/(-1805042263882410*u-189114296805*u^2+1594323*u^4+5245736013*u^3+406272576213372225)

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This seems interesting, but could you clarify? You have shown that C is of genus 1 (birationally, presumably f=0 is quite singular). And that C has a point of infinite order? –  Charles Matthews Aug 2 '13 at 12:43
    
@CharlesMatthews edited the question giving 8 points of infinite order and a map from the Weiersstras model to C. –  joro Aug 2 '13 at 13:09
    
I suspect gcd(x+z,y-z)<max(|x+z|,|y-z|)^E for E small enough might be a putative counterexample to abc for integers, though this is extremely unlikely. There were some large abc triples of moderate quality and low merit after clearing the gcd. –  joro Aug 2 '13 at 16:51
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3 Answers 3

There is an "abc theorem for higher genus curves".

Let $a,b,c$ be three sections of a line bundle of degree $d$ on a curve $C$ of genus $g$ with no common zeroes such that $a+b=c$. Then the number of zeroes of $abc$ is at least $d+2-2g$.

Proof: Let $f = -a/b$ be a map $C \to \mathbb P^1$ of degree $d$. By Riemann-Hurwitz, the total ramification degree is at most $2d+2g-2$. But the total ramification is at least the ramification over the points $0$, $1$, and $\infty$, which is equal to $3d$ minus the number of zeroes of $abc$.

We conclude that if $x'$, $y'$, and $z'$, the pullback of $x+z$, $y-z$, $-3x+4y$ to $E$ are three sections such that $x'^5 + y'^5= z'^2 (x'+y')^3$, then $x'$ and $y'$ cannot be relatively prime (and if we try harder, we can compute how many factors they must share). This is vindicated by the fact that $(x',y',z')=(0,0,1)$ or $(x,y,z)=(-1,1,1)$ is a point on the curve, and is in fact the singular point, so its preimage on $E$ has large degree.

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Thanks. This doesn't appear to be unconditional as the second question asks? –  joro Aug 2 '13 at 15:15
    
It is, sorry for the confusion. –  Will Sawin Aug 2 '13 at 15:17
    
I don't understand algebraic geometry, but I fail to see why gcd(x',y') should be large. Would you please try non-rigorous down to earth explanation? Thanks. –  joro Aug 2 '13 at 16:09
    
I suspect gcd(x',y')<max(|x'|,|y'|)^E for E small enough might be a putative counterexample to abc for integers, do you have opinion about this? –  joro Aug 2 '13 at 16:47
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Historical remark: Mason and Stothers stated their result for arbitrary curves, not just $\mathbb{P}^1$. It is later expository accounts that consider just polynomials. –  Felipe Voloch Aug 2 '13 at 20:31
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The big gcd can be explained as follows: Upon replacing $x$ with $x-z$ and $y$ with $y+z$, we obtain the slightly more convenient equation \begin{equation*} x^5+y^5=(-3x+4y+7z)^2(x+y)^3. \end{equation*}

On the other hand, we have the identity \begin{equation*} x^5+y^5=(x^2 - 3xy + 6y^2)(x+y)^3-5y^3(y+x)(y+2x), \end{equation*} so if $(x,y,z)$ is a point on the curve, then $(x+y)^3$ divides $5y^3(y+x)(y+2x)$, hence

\begin{equation*} (x+y)^2\text{ divides }5y^3(y+2x). \end{equation*}

By homogeneity we may assume that $\text{gcd}(x,y,z)=1$. Set $d=\text{gcd}(x,y)>0$. Now $(x+y)/d$ is relatively prime to $y/d$ and $(y+2x)/d$. So the previous divisibility yields that $(\frac{x+y}{d})^2$ divides $5d^2$. In particular \begin{equation*} d\ge \frac{\lvert x+y\rvert^{1/2}}{5^{1/4}}. \end{equation*} I believe that a finer kind of this reasoning can improve the lower bound for $d$.

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$(x+y)^2\text{ divides }5y^3(y+2x)$ certainly doesn't imply gcd(x,y) > 1. There are a lot of counterexamples found by fast bruteforce search. Especially when |x+y|=1. –  joro Aug 6 '13 at 10:46
    
@joro: I didn't say that $gcd(x,y)>1$. All what I say is that $gcd(x,y)$ grows in terms of $\lvert x+y\rvert$. Actually, for any fixed $x+y$, that is setting $y=a-x$ for a fixed integer $a$. you get an elliptic curve in $x$ and $z$, which then has only finitely many integral points. –  Peter Mueller Aug 6 '13 at 14:33
    
I see. It is still not entirely clear how big |x+y| is. In your example with $y=a-x$ there are infinitely many choices for $a$, so either the curves are birationally equivalent or still $x,z$ can't be sufficiently larger than $a$ ? –  joro Aug 6 '13 at 14:50
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The situation appears more complicated.

$abc$ implies something stronger than just large gcd.

The identities are of the form $$ a^5 + b^5 = c^2 d^3$$

Set $g^5 = \gcd(a^5,b^5)$. If $g^2 || c^2, g^3 || d^3$ we still will have sum of two fifth powers being squarefull, which contradicts $abc$.

So abc implies sufficiently large squarefree part of $c^2 d^3$ after clearing the gcd.

Experimentally, $|7^3 d^3|=g^6$, which gives identities of the form

$$ a'^5 + b'^5 = c'^2 d' \qquad (1)$$

(1) gives $abc$ triples of barely good quality.

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You seek an explanation for $|7^3d^3|=|g^6|$. I believe I can provide one. Recall that $d=a+b$, and that this is our equation. $a^5+b^5=c^2(a+b)^3$. The equation for a singular form of the elliptic curve is gotten by dividing both sides by $(a+b)$. The singular point is $a=b=0$, $c=1$. It has two inverse images on the elliptic curve, and the order of vanishing of $a$ and $b$ at those images is $1$, while the order of vanishing of $a+b$ is $2$. Since $g$ comes from these points, we expect that $g^2$ divides $d$, which implies the empirical result. –  Will Sawin Aug 19 '13 at 21:32
    
Mathematically, note that $(a+b)^4+5(a-b)^4 +10 (a+b)^2(a-b)^2 = c^2 (a+b)^2$, hence $(a+b)^2$ divides $(a-b)^4$ (on the curve), hence $(a+b)$ divides $(a-b)^2$ mathematically. We conclude that $d$ divides $4g^2$. –  Will Sawin Aug 19 '13 at 21:37
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