# Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z[t]$ be unbounded as n varies?

We've recently seen this question: Can the number of solutions $ab(a+b+1)=n$ for $a,b,n \in \mathbb{Z}$ be unbounded as $n$ varies? It appears initially plausible that the answer is yes, but evidently there are good reasons to believe that the answer might be no. The same question and lack of certainty applies equally to $$ab(a+b-1)=n.$$

I wonder about the analogous situation with $\mathbb{Z}$ replaced by $\mathbb{Z}[t]$

Is there a known or conjectured upper bound for the number of solutions $ab(a+b-1)=n$ for $a,b,n \in \mathbb{Z}[t]$ as $n$ varies? What is the best known lower bound?

Meta-reasoning tells us that the number of solutions for this $\mathbb{Z}[t]$ problem is not know or suspected to be unbounded because that would supply us with parametric families of solutions to the problem in $\mathbb{Z}$.

I picked the $-$ version because I was able to find this nice example:

$$[a,b,a+b-1]=[1,2t(t+1)(2t+1),2t(t+1)(2t+1)],[2t^2,2(t+1)^2,(2t+1)^2]$$ and $$[t(2t+1),(t+1)(2t+1),4t(t+1)]$$ are three solutions to $$ab(a+b-1)=\left(2t(t+1)(2t+1)\right)^2.$$ That counts as 3 solutions. (or 6 or 18 if we take $[b,a,a+b-1]$ and $[1-a-b,a,-b]$ as different, though why bother?). Looking for integer solutions to $ab(a+b-1)=\left(2t(t+1)(2t+1)\right)^2\left(2t(t+1)(2t+1)\right)^2$ most often leads to just the three anticipated solutions but sometimes there are more. Including 6 for $t=55$ and 9 for $t=175$.

It seems reasonable to require that the leading coefficients be positive and that $n=n(t)$ not be an integer (I needed to say that before someone else did!).

The more general question arises as well. But I like concrete examples and don't want to venture a version of minimal Weierstrass form (feel free to enlighten me however)

I've further extended the previous update to this question and made it an answer. Please provide a better one!

-
Aaron, I understand why you restrict you considerations to $\mathbb Z[t]$ but $\mathbb Q[t]$ (or even $\overline{\mathbb Q}[t]$) are more natural. Even this sounds too cosmetic, it will make the equation more general, as $a+b-1$ may be replaced with $a+b-\text{const}$ by appropriate scaling. Also, your solution will be "nicer" after shifting $t$ by half. – Wadim Zudilin Jan 5 '11 at 13:44
Wadim, $\mathbb{Q}[t]$ might give some general insight (I'll think about it) but I am mainly interested in the implications for $\mathbb{Z}$. That is why I don't want to scale for this problem. I suppose that the group addition will lead to rational functions in $t$. If one substitutes $t+1/2$ for t then two of the solutions still have integer coefficients but the middle one does not. Is that what you meant? – Aaron Meyerowitz Jan 5 '11 at 16:08
Would $\mathbb{Z}[t,u]$ make sense? – jerr18 Jan 6 '11 at 8:38
Definitely. However if one has $f(t,u) \in \mathbb[t,u]$ then $g(t)=f(t^2-43,t+17)$ etc. should give a plethora of one variable things to find. – Aaron Meyerowitz Jan 6 '11 at 8:55
FYI I am unsuccessfully trying to equate coefficients assuming deg(y)=1 – jerr18 Jan 6 '11 at 9:37

Let $F(x,y)$ be any degree 3 polynomial. So for example, you can take $F(x,y)=xy(x\pm y\pm 1)$. The original question asked about solutions to $F(x,y)=n$ with $x,y \in \textbf{Z}$, and the current question replaces this with $x,y \in \textbf{Z}[t]$. More generally, let $n\in\textbf{C}[t]$ be a nonconstant polynomial, let $I(n)$ be the number of solutions to $F(x,y)=n$ with $x,y \in \textbf{C}[t]$, and let $R(n)$ be the rank of the group of points on the elliptic curve $E_n : F(x,y)=n$ with $x,y\in\textbf{C}(t)$.

Theorem: If $E_n$ is an elliptic curve and if the $j$-invariant of $E_n$ is not in $\textbf{C}$, then there is an absolute constant $K$ such that $I(n) \le K^{1+R(n)}$.

This is a result of Marc Hindry and mine. (We actually only prove it for standard Weierstrass equations, but the proof is the same in general.) The reference is Invent. Math. 93 (1988), 419-450, although it also relies heavily on J. Reine Angew. Math. 378 (1987), 60-100. If $E_n$ has constant $j$, there's a similar statement, but one may need to untwist first. For example, taking $F(x,y)=xy(x+y)$, one needs to restrict to $n$'s that are cube-free. Also, of course, one has to avoid $F$ and $n$ values such that the curve $F=n$ has a singularity, since then $E_n$ is no longer an elliptic curve.

All of this doesn't answer the question, but it suggests that it will be hard to make $I(n)$ large, since it's not even known that $R(n)$ can be arbitrarily large.

-
Wonderful. I'll check that out. Does your paper address the case that one assumes say $2t^2-1$ is a square as I did above? One can probably only get that boost once. I don't expect that $I(n)$ can be arbitrarily large (since there are doubts about that for $\mathbb{Z}$) – Aaron Meyerowitz Jan 11 '11 at 2:54
If you assume that $2t^2-1=s^2$, you're working over the function field of that curve. Our paper deals with the general case of an elliptic curve defined over the function field of an arbitrary curve. So yes, our paper would include the case that $2t^2-1=s^2$. (BTW, it's possible that taking $2t^2-1=s^2$ increases the rank.) In general, if the base curve has genus $g$, then the constant $K$ in my response will depend on $g$, since the bound for the (easy to prove) function field $abc$-conjecture depends on $g$. – Joe Silverman Jan 11 '11 at 12:34
Please excuse my dumbness. The integral points on $(xy+a^2)(x+y+a)=ba^3 \; (1)$ puzzle me. I suppose they can be arbitrary large by scaling rational points on $(xy+1)(x+y+1)=b$ by $a$. If I have done the computations right, the minimal model of $(1)$ depends on both $a$ and $b$. BTW, in your answer can the coefficients of $F(x,y)$ be in $C(t)$? – jerr18 Jan 13 '11 at 13:09
@jerr18: Yes, you can take $F(x,y) \in \mathbf{C}(t)[x,y]$. – Joe Silverman Jan 13 '11 at 16:23
I am missing something. Aaron's example can be scaled to arbitrary many solutions in $\mathbb{Z}[t]$ to $xy(x+y-a)=na^3$ via scaling rational points on $xy(x+y-1)=n$ by $a$. I would expect the $j$ invariant to depend on $n$ which is non-constant. – jerr18 Jan 14 '11 at 11:19

The $\mathbb{Z}[t]$ example above with 3 (pairs of) positive solutions for $ab(a+b-1)=n$ has 5 in $\mathbb{Z}[u,v]/(u^2-2v^2+1)$ if $t=\frac{(2v+u)(u-v)}{2}$. This gives us a parametric family of curves in $ab(a+b-1)=n$ in $\mathbb{Z}$ with 5 pairs of positive integer points . We need $u>v$ so first cases are $(u,v,n)=(7,5,21420^2),(41,29,840456540^2),(99,70,32947602728040^2)$. In the first case $n=21420^2$ there is a further sixth pair of solutions but not for the next the next 4 cases.

Suppose that $u^2=2v^2-1$ and let $t=\frac{(2v+u)(u-v)}{2}$. Then we have the same three solutions as before for $ab(a+b-1)=(2t(t+1)(2t+1))^2$. There are also two more. (plus the same things with $a$ and $b$ swapped). These are as follows with the second and third being the new points:

$[{\small 1}, \frac{(2v+u)(2v-u)(u+v)(u-v)uv}{2},\frac{(2v+u)(2v-u)(u+v)(u-v)uv}{2}]$

$[\frac{(u-v)^2(2v+u)u}{2},{\small (2v-u)(2v+u)v^2}, \frac{(2v-u)(u+v)^2u}{2}]$

$[\frac{(2v-u)^2(u+v)v}{2}, {\small(u-v)(u+v)u^2}, \frac{(u-v)(2v+u)^2v}{2}]$

$[\frac{(2v+u)^2(u-v)^2}{2}, \frac{(2v-u)^2(u+v)^2}{2}, { \small u^2v^2}]$

$[\frac{(2v+u)(u-v)uv}{2}, \frac{(2v-u)(u+v)uv}{2}, {\small (2v-u)(2v+u)(u-v)(u+v)}]$

note that $\frac{2v-u}{u-v},\frac{v}{u},\frac{2v+u}{v+u}$ are three successive convergents to $\sqrt{2}$. Of course these can be rewritten using the relation $u^2=2v^2-1$ and/or substitutions, for example the product $(2v-u)(2v+u)=1$.

In another direction is again 3 points but in $\mathbb{Z}[t,x]$

${\small [1+(4t+2)x, 2(t+1)(2t+1)(t+(4t^2+2t-1)x), 2t(2t+1)(t+1+(4t^2+6t-1)x)]}$

${\small [2t(t+(4t^2+2t-1)x), 2(t+1)(t+1+(4t^2+6t-1)x), (2t+1)^2(1+(4t+2)x)]}$

${\small [(2t+1)(t+(4t^2+2t-1)x), (2t+1)(t+1+(4t^2+6t-1)x), 4t(t+1)(1+(4t+2)x)]}$

For a majority of values for $t,x$ the resulting curve over $\mathbb{Z}$ has three positive points. Hower cases of 4 and 5 do happen quite frequently and even 6 sometimes. This suggests that more is there to be found. Replacing $x$ with $-x$ and then multiplying through by $-1$ gives curves $ab(a+b+1)=n$

The first example represents two iterations of a process:

1. Find a family of examples in $\mathbb{Z}$ (say look at ones with at least 3 solutions, one being $[1,b,b]$, from those notice that a few have the other two triples very close to each other.

2. Find a parametric form in $\mathbb{Z}[t]$.

3. Put in integers for $t$ to get curves over $\mathbb{Z}$ with at least three points and look for cases with higher rank.

4. Note that several have 5 solutions and prime factors familiar from convergents to $\sqrt{2}.$

5. Find a parametric form of that.

The second example follows a similar process but this is long enough.

-
Do you think it is possible over $\mathbb{Z}[t,x]$ to exist many $n(t)$ with more than one solution with $deg(a)=deg(b)=1$? If so, equating $a,b$ for the $n$s probably can be done with linear algebra (probably at the cost of going to $\mathbb{Q}$ but without using the group law). Bugs me if possible the degree will be low, though. – jerr18 Jan 13 '11 at 14:27
I'm not sure exactly what you mean. Wouldn't it be $n(t,x)$? Do you mean the degree in $t$ or in $x$? Ultimately I want to find out things about integer curves with many points. I'd expect better results from looking at something like the nth 2nth and 3nth convergents to $\sqrt{p}$ (say for $p=2$) but the integers are huge making them hard to investigate numerically and that might only go from 5 to 6 or 7 integer points (if it works at all). – Aaron Meyerowitz Jan 13 '11 at 16:39
Sorry, I meant $n(t,x)$. By deg(1) I mean both a,b to be of linear of form $c_t t + c_x x + c_0$ and $c_t, c_x, c_0 \in Z$ – jerr18 Jan 13 '11 at 16:58
And if linear a,b are possible my hope is to equate one of the solutions with linear algebra in the hope that the other solutions will differ. – jerr18 Jan 13 '11 at 17:02
Hm, since you accepted this answer does it mean more solutions aren't possible? – jerr18 Jan 27 '11 at 8:18

This is not an answer, just a long comment.

Out of opportunism I tried multiplying Aaron's record by a square. $(2t+3)^2$ turned somewhat lucky.

$n$ with 3 solutions: $2^{2} \cdot t^{2} \cdot (t + 1)^{2} \cdot (2t + 1)^{2} \cdot (2t + 3)^{2}$

Solutions: $$\left[2^{2} \cdot t \cdot (t + 1), (t + 1) \cdot (2t + 1)^{2}, t \cdot (2t + 3)^{2}\right]$$ $$\left[2 \cdot (2t + 3) \cdot t^{2}, (2t + 1) \cdot (2t + 3), 2 \cdot (2t + 1) \cdot (t + 1)^{2}\right]$$ $$\left[2 \cdot t \cdot (t + 1) \cdot (2t + 1) \cdot (2t + 3), 1, 2 \cdot t \cdot (t + 1) \cdot (2t + 1) \cdot (2t + 3)\right]$$

Another $n$ with 3 solutions is $t^{2} \cdot (t + 2)^{2} \cdot (t + 1)^{4}$

Later, using a symbolic approach found these 5 $n$ with 3 solutions: $[t^{8} + 8t^{7} + 26t^{6} + 44t^{5} + 41t^{4} + 20t^{3} + 4t^{2}, 1024t^{8} + 4096t^{7} + 6528t^{6} + 5248t^{5} + 2212t^{4} + 456t^{3} + 36t^{2}, 256t^{8} + 1024t^{7} + 1664t^{6} + 1408t^{5} + 656t^{4} + 160t^{3} + 16t^{2}, 64t^{8} + 384t^{7} + 928t^{6} + 1152t^{5} + 772t^{4} + 264t^{3} + 36t^{2}, 64t^{8} + 384t^{7} + 928t^{6} + 1152t^{5} + 772t^{4} + 264t^{3} + 36t^{2}]$

-
So this is two other cases with 3 solutions. Nice. I think I can get to 5 solutions for something like $t=uv$ where $2v^2=u^2+1$ but the details need to be pinned down. – Aaron Meyerowitz Jan 8 '11 at 22:24
I suspect doing intersections better be done with 2 solutions and lower degree... – jerr18 Jan 9 '11 at 8:03
$(t + 2)^{2} \cdot (2t + 1)^{2} \cdot t^{4}$ has 2 solutions. – jerr18 Jan 9 '11 at 16:29
$2^{2} \cdot 3^{2} \cdot t^{2} \cdot (2t + 1)^{2} \cdot (3t + 1)^{2}$ has 3 solutions. – jerr18 Jan 9 '11 at 17:18