# Examples of Diophantine equations with a large finite number of solutions

I wonder, if there are examples of Diophantine equations (or systems of such equations) with integer coefficients fitting on a few lines that have been proven to have a finite, but really huge number of solutions?

Are there ones with so large number of solutions that we cannot write any explicit upper bound for this number using Conway chained arrow notation?

Update: I am also interested in equations with few solutions but where a value in a solution is very large itself.

• You might want to enforce some restriction on the number of unknowns in the Diophantine equation, else, we can have some trivial equations like $\displaystyle \sum_{k=1}^n x_k^2 = 1$, where $n$ is large, which has $2n$ solutions.
– user17762
May 10, 2013 at 21:16
• @user17762 The explicit form of the equation that do not use abbreviations like $\sum, \prod$ should fit on a few lines. May 10, 2013 at 21:21
• For the question of very large solutions, you may find something at the Mathoverflow question on "eventual counterexamples". Jun 8, 2013 at 4:47
• Archimedes' famous "Cattle of the Sun" problem has a single solution, with 206,544 digits ... but, since it's famous, you probably already know about it.
– Blue
Jun 8, 2013 at 5:12

I think the rough answer to your question is that "there are no such natural equations." Let me try to justify this heuristic.

Caporaso, Harris, Mazur; and Baker's method: Consider diophantine equations corresponding to curves, for example equations of the form:

$$C: f(x,y) = 0$$

where $f(x,y)$ is a polynomial with integer coefficients. One knows that the number of solutions (with $x$ and $y$ in $\mathbf{Z}$ or $\mathbf{Q}$) is --- to some extent --- determined by the complex geometry of $C$. If the genus $g$ of $C$ is greater than two, then Faltings proved that $C$ has only finitely many rational points. However, much more (may) be true. Assuming a conjecture of Lang, it was proved by Caporaso, Harris, and Mazur (J. Amer. Math. Soc. 10 (1997), 1-35) that the number of rational points $\#C(\mathbf{Q})$ is bounded by a function $A(g)$ which only depends on $g$, not on $C$. This doesn't prove that $A(2)$ [for example] is not enormous, but all known lower bounds in $g \ge 2$ on the number of rational points are at most polynomial. For example, I think the largest number of rational points on a genus two curve that has ever been found has at most a few hundred rational points. Since the genus is controlled by the degree of $f$, this suggests that it is highly unlikely to find an equation with small coefficients and degree which has absolutely massive solutions, even if one makes the coefficients big. Note that this is only a heuristic, but it strongly suggests that there are no such equations. It certainly says that no such equations are currently known.

One is in slightly better shape (or worse, depending on what you're trying to do) if one restricts to integral points of $C$. For certain classes of equations (say $F(x,y) = m$ for homogenous $F$) there are explicit bounds on the number of integral solutions in terms of the coefficients coming from Baker's theorem (on linear forms in logarithms). The bounds are not great (perhaps super-super exponential in the coefficients), but they certainly preclude numbers of the size you are interested arising from anything that one can sensibly write down. Moreover, one conjecturally expects that Baker's bounds are not optimal.

For genus 1 and 0, the situation is similar. There's difference here again between whether one wants to consider only integral solutions or rational solutions, but the result (in either case) is that if one insists that there are only finitely many solutions (in either integers or rationals), then there is a bound for the largest such solution in terms of the coefficients (which, using Baker's method for integral points, might be quite large, but is still tiny compared to the numbers you are discussing).

Moral: If one assumes Lang's conjecture (and we certainly don't have any evidence that it is false), there is still heuristic evidence to suggest that the the types of equations you are looking for do not exist, either for curves or higher dimensional varieties: this is very speculative, but certainly means that writing down any such equations is either impossible or very hard.

Exponential Diophantine Equations: There are other flavours of Diophantine equations besides algebraic varieties. For example, you could consider exponential Diophantine equations, e.g.: $2^n = x^2 + 23$. In this case, one can often convert solutions to these equations into subsets of polynomial equations of genus at least two. For example, the previous equation gives solutions to one of the five genus three curves: $$Ay^5 = x^2 + 23, \qquad A = 1,2,4,8,16.$$

Thus one can often "reduce" to the case of curves. Alternatively, Baker's method can often be used directly on exponential equations to give upper bounds.

The phenomena of a big "smallest" solution: There are some equations which have infinitely many solutions of which the "smallest" is quite large, for example, the first non-trivial solution to Pell's equation $x^2 - 61 y^2 = 1$ has $$[x,y] = [1766319049, 226153980].$$ However, in these cases (and in similar cases coming from elliptic curves) one at least expects (and knows for Pell) that there exists a bound on the regulator which is polynomial in the coefficients, where the regulator involves the logarithm of the entries. So this gives upper bounds for primitive solutions which are exponential in a polynomial in the coefficients, still enough to prevent super-huge numbers arising as the smallest interesting solution.

Summary: There are many other flavours of Diophantine equation one can write down, but I think the summary is that the conjectures of Lang suggest, on some heuristic level, that the type of phenomena you are looking for simply does not occur naturally. There may be a way of embedding Ackerman's function in to some system of equations (but not into polynomial equations), but I think you would consider that cheating.

• The phenomena of a big "smallest" solution $-$ very interesting for me. Know you any links/literature related to? Jun 30, 2013 at 23:58

While I largely agree with @Qbert's sentiment that you won't find anything that's hard to bound with Conway chained arrow notation (note that Ackermann's function and Graham's number only need 2 or 3 arrows, so even allowing the shortcut "where G is Graham's number" won't get you to the next line), here is something explicit that at least you might want to use two arrows for. $$a^3-k^2=-11492\\ x^2-(a^2-1)y^2=1\\ u^2-(a^2-1)v^2=1\\ s^2-(b^2-1)t^2=1\\ v=ry^2 \quad b=1+4py \quad b=a+qu \\ s=x+cu \quad y=k+e-1 \\ t=k+4(d-1)y\\ x = w+z \\ f^3-g^2=w$$

I consider solutions in positive integers. (You can add constraints like $w=1+A^2+B^2+C^2+D^2$ where necessary to force variables to be positive if you prefer to work over all of $\mathbb Z$.)

The first equation has a solution with $a=154319269$ as given by Elkies here (and also others, e.g. $a=13,k=117$). The next six lines are exactly the system given by Martin Davis for Theorem 3.1 in "Hilbert's Tenth Problem is Unsolved" and only have one solution for given $a,k$ when $x$ is exactly the $k$th smallest solution to $x^2-(a^2-1)y^2=1$, in which case $x\ge a^k$. The next lets $w$ range over all $1\le w < x$.

The first and last equations are Mordell curves and have only finitely many solutions for $w$ fixed. Taking the value for $a$ given above, there is a solution with $x>10^{10^{13}}$. Then I couldn't really say how many solutions there might be to the last equation, but the largest is at least $\lfloor\sqrt{x-2}\rfloor$, and based on Marshall Hall's conjecture we might expect it to be greater than $x^6$.

But in any case a bound from Baker given in section V.8 of Silverman & Tate would require $$g\le\exp((10^6w)^{10^6})<3^{10^{10^{20}}}<10\rightarrow 5 \rightarrow 2$$ using Conway's notation (and assuming the $a$ given above is the largest solution to the first equation).

I couldn't give you an explicit equation, but if you're willing to bend the rules a bit, I'm fairly certain that one should exist. Apply Matiyasevich's theorem to the busy beaver problem: for any $n$, the set of running times of all halting binary Turing machines with $n$ states is finite and Diophantine, and therefore represented by the positive values taken by some integer polynomial $P(x_1,\ldots,x_k)$.

Then one could choose the Diophantine equation $P(x_1,\ldots,x_k) = 1+a^2+b^2+c^2+d^2$, with the proviso that there could be infinitely many solutions (though perhaps someone familiar with the MRDP construction could correct me), but there are only finitely many distinct values of $(a,b,c,d)$. And the busy beaver function grows absurdly (non-computably) fast, so it shouldn't take very long for the values of $a$, $b$, $c$, $d$ to outstrip anything that arrow notation can describe.

• Although you did not give an explicit equation, your argument seems convincing. Matiyasevich theorem indeed establishes that any computable set is a projection of the set of solutions of some Diophantine equation. So, basically, we have a way to encode any algorithm (in particular, one that computes expressions given in Conway chained arrow notation, or even more fast-growing functions) using a Diophantine equation. Based on examples of universal Diophantine equations, I am inclined to believe it can fit on several lines. So, the bounty is yours! Thanks! Jun 12, 2013 at 2:08
• The basic idea is sound and gives a good intuition. However it might be useful to note that the number of distinct runtimes for machines of size (exactly) $n$ isn't that large (to a CS guy like me at least). In fact it is something like $O(2^n)$, since there are about that many distinct Turing Machines of size $n$ (and surely there cannot be more finite runtimes than there are machines!). Some of the runtimes are huge, but the size of the set itself is not absurd.
– cody
Mar 16, 2015 at 19:55
• An easy fix would be: the set of numbers smaller than the max runtime of a machine of size $n$. That set is certainly r.e., and so Diophantine by Matiyasevich's theorem.
– cody
Mar 16, 2015 at 19:58
• Another way to adjust this is to note that the number of representations of $n$ is $\Omega(n)$ provided $n$ is odd. So the size of the solutions on the left (suitably adjusted to be odd) directly translates to having many solutions on the right. Aug 21, 2019 at 20:40

There are many such equations. Here is one such equation.

Note that the number of ways of writing a number $n$ as a sum of $2$ squares, where $$n=2^{a_0}\left(p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k} \right)^2 q_1^{b_1} \cdots q_s^{b_s}$$ $p_i \equiv 3\pmod4$, $q_j \equiv 1 \pmod4$ and $p_i,q_j$ are primes, is given by $$f(n) = 4B$$ where $B = (b_1+1)(b_2+1)\cdots(b_s+1)$. Choose a large enough $B$. Construct a number $n$ with the given $B$ (You have plenty of options of creating $n$). Form the diophantine equation $x^2+y^2 = n$. This has $4B$ solutions.

You can create similar examples like $x^2 + 2y^2 = n$, $x^2 + 3y^2 = n$ and letting $n$ to be large with some constraints, you can get as many number of integer solutions, you want.

• Perhaps even simpler is $x^2-y^2=n$. For odd $n$, the number of solutions is half the number of divisors of $n$. May 11, 2013 at 0:17
• To increase the number of solutions, I have to increase n, and I can only do this while its decimal representation fits in a few lines. The max number of solutions I can get this way does not look very large. I suppose, it is easy to write a larger number using Conway chained arrow notation. May 11, 2013 at 1:55
• @Vladimir I would guess defining what you mean by "explicit form" would be tricky. For instance, would you call $\pi$ to be an explicit form?
– user17762
May 11, 2013 at 2:04
• @GerryMyerson Yes. It is interesting that my mind skipped the easier one and went for a slightly harder one.
– user17762
May 11, 2013 at 2:12
• @user17762 I mean polynomial Diophantine equations with integer coefficients, all integers are written in decimal representation. May 11, 2013 at 2:56

The smallest solution to the system $$x^2+101y^2=z^2,\quad x^2-101y^2=t^2$$ is $$x=2,015,242,462,949,760,001,961;\quad y=118,171,431,852,779,451,900;$$ $$z=2,339,148,435,306,225,006,961;\quad t=1,628,124,370,727,269,996,961$$ according to Wolfram.

Here is simpler example with large solution.

Fix positive integers $a>1,n>2$.

The hyperelliptic curve $y^2=x^{2n}+x-a$ has finite number of rational solutions.

One integer solution is $(a,a^n)$.

All functions g:N-->N which have a single-fold Diophantine representation are dominated by a limit-computable function f:N{0}-->N which is implemented in MuPAD and whose computability is an open problem

http://arxiv.org/abs/1309.2682

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– Joao
Oct 17, 2014 at 7:26

Let $b$ is a non-zero integer, and let $n$ is a positive integer. The equation $y(x-b)=x^n$ has only finitely many integer solutions. The first solution is: $x=b+b^n$ and $y=(1+b^{n-1})^n$. The second solution is: $x=b-b^n$ and $y=-(1-b^{n-1})^n$,

cf. [1, page 7, Theorem 9] and [2, page 709, Theorem 2].

The number of integer solutions to $(y(x-2)-x^{\textstyle 2^n})^2+(x^{\textstyle 2^n}-s^2-t^2-u^2)^2=0$ grows quickly with $n$, see .

References

 A. Tyszka, A conjecture on rational arithmetic which allows us to compute an upper bound for the heights of rational solutions of a Diophantine equation with a finite number of solutions,

http://arxiv.org/abs/1511.06689

 A. Tyszka, A hypothetical way to compute an upper bound for the heights of solutions of a Diophantine equation with a finite number of solutions, Proceedings of the 2015 Federated Conference on Computer Science and Information Systems (eds. M. Ganzha, L. Maciaszek, M. Paprzycki), Annals of Computer Science and Information Systems, vol. 5, 709-716, IEEE Computer Society Press, 2015.

 A. Tyszka, On systems of Diophantine equations with a large number of integer solutions,

http://arxiv.org/abs/1511.04004