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Given a polynomial $P$ with integer coefficients in finitely many variables, we denote by $v(P)$ the product of the absolute values of the non-zero coefficients and the non-zero total degrees of the monomials of $P$.

Since there is no general algorithm to decide whether an equation $P(x_1, \dots, x_k) = 0$ has a solution in integers $x_1, \dots, x_k$ or not, for every computable function $f$ there is a polynomial $P$ such that the equation $P(x_1, \dots, x_k) = 0$ has such solution, but none for which all $x_i$ have absolute value less than $f(v(P))$. In particular, this holds for the function $f: n \mapsto 2^n$.

For the purposes of this question, we say that the diophantine equation $P(x_1, \dots, x_k) = 0$ has large smallest solution if it has a solution, but none such that all $x_i$ have absolute value less than $2^{v(P)}$.

Question: Which are ''nice'' examples of diophantine equations with large smallest solution?

Here are some non-examples:

  • Put $P := x-1$. Then, $v(P) = 1$, but the equation $P(x) = 0$ has the solution $x = 1$ whose absolute value is less than $2^{v(P)} = 2^1 = 2$.

  • Put $P := a^2 + b^2 - c^2$. Then, $v(P) = 2^3 = 8$, hence our bound is $2^8 = 256$. Even not counting the solution $a = b = c = 0$, there is e.g. the solution $a = 3, b = 4, c = 5$ where all values are less than $256$.

  • Put $P := x^2 - 61y^2 - 1$. Then, $v(P) = 2^2 \cdot 61 = 244$, and the solution $(x,y) = (1766319049,226153980)$ is clearly less than the bound of $2^{244} > 10^{73}$.

  • Put $P := a^4 + b^4 + c^4 - d^4$. Then, $v(P) = 4^4 = 256$, hence our bound is $2^{256}$. Even not counting the solution $a = b = c = d = 0$, there is e.g. the solution $a = 95800$, $b = 217519$, $c = 414560$, $d = 422481$, where all values are less than $2^{256} > 10^{77}$.

  • Put $P := x^3 + y^3 + z^3 - 33$. Then, $v(P) = 3^3 \cdot 33 = 891$, and the absolute values of all numbers in the solution $$ (x,y,z) = (−2736111468807040,−8778405442862239,8866128975287528) $$ are clearly less than the bound of $2^{891} > 10^{268}$.

  • Now, let's have a look at the equation $$ \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \ = \ 4. $$ This truely has a large smallest solution, at least if one insists on $a$, $b$ and $c$ being positive, hasn't it? —

    No, not really: subtracting 4 from both sides of the equation, and multiplying the equation by the common denominator of the terms, we arrive at the polynomial $$ P = a^3-3a^2b-3a^2c-3ab^2-5abc-3ac^2+b^3-3b^2c-3bc^2+c^3, $$ which has $$ v(P) = 3 \cdot 3^2 \cdot 3^2 \cdot 3^2 \cdot 3 \cdot 5 \cdot 3^2 \cdot 3 \cdot 3^2 \cdot 3^2 \cdot 3 = 215233605. $$ We observe that the smallest solution even doesn't exceed $2^{267}$, let alone $2^{215233605}$.

Addendum: In order to further explain what I mean by ''nice'' in this context: while it is straightforward to build equations which in some obvious ways encode computations which yield large numbers, the art is rather to find equations which do not do so in any obvious way (like the non-examples above), but which still satisfy the condition on the size of the smallest solution. — And these latter equations are rather the ones which are ''nice'' in the sense of this question. Nevertheless, in case no good such examples are found, also the equations of the former type are valid answers to the question (just not such good ones).

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  • $\begingroup$ Is the intention in the question to allow equations where we consider only positive solutions or only nonzero solutions, as you do in some of the examples? $\endgroup$
    – Will Sawin
    Commented Mar 20 at 15:54
  • $\begingroup$ @WillSawin Nice examples are welcome in any case. Just the smallest solution (respectively, smallest non-zero or smallest positive solution, if applicable) should exceed the given bound. $\endgroup$
    – Stefan Kohl
    Commented Mar 20 at 16:02
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    $\begingroup$ @DenisShatrov If I counted right that has $v(P)= 324$ but the solution in that question is merely of size $\approx 2^{140}$, so not large enough. $\endgroup$
    – Will Sawin
    Commented Mar 20 at 19:12
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    $\begingroup$ $(x_1-2)^2 + \sum_{i=1}^{k-1} (x_i^2 - x_{i+1})^2$ doesn't meet your requirements, but arguably it gets somewhat "close". The smallest solution has size $2^{2^k}$, and $v(P) = 32 \times 48^{k-1}$. $\endgroup$
    – D.W.
    Commented Mar 21 at 1:46
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    $\begingroup$ Some very large numbers come up in smallest solutions to problems in the thread math.stackexchange.com/questions/514/… I don't know what those problems would look like if you turned them into diophantine equations. Likewise, mathoverflow.net/questions/15444/… $\endgroup$ Commented Mar 21 at 2:32

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A better example: I believe conjecturally the equation $(a-N)^2 + (b-a^2)^2 + (c-b^2)^2 + (d-c^2)^2 + (x^2 - (d+a-1) y^2+1)^2 $ in the variables $a,b,c,d,x,y$, with $N$ fixed, should do the trick for most values of $N$.

We have $v(P) =2^{29} 3^8 5^4 N^3$.

The first few terms force $d= N^8$ and the last term gives the negative Pell's equation $x^2- (N^8+N- 1)y^2=-1$. This is solvable for a majority of integers by work of Koymans and Pagano On Stevenson's Conjecture and there is no reason to suspect the proportion is much different for integers of the form $N^8+N-1$ (in particular, since $N^8+N-1$ is an irreducible polynomial).

The least solution of Pell's equation is of size roughly exp of the square root of the discriminant as long as the class group is not too large, which it is under the natural restriction of the Cohen-Lenstra conjecture to a polynomial subsequence of discriminants.

So you just have to take $N$ large enough that exp of $N$ to the fourth power beats exp of $N$ to the third power times a rather large constant, so $N$ very roughly on the order of $2^{29} 3^8 5^4 \approx 2,000,000,000,000,000$ should work.

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  • $\begingroup$ Well — but doesn't the Pell equation in this always have the solution $(x,y) = (N^4,1)$, or did I misread something? $\endgroup$
    – Stefan Kohl
    Commented Mar 20 at 17:59
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    $\begingroup$ @StefanKohl Right, true. Sufficiently many minor adjustments should make that disappear, e.g. This one with $N^8-1$ might be OK. One might need to take $N^8 + N + 1$ or something like that, which is a much nicer polynomial for this sort of thing. $\endgroup$
    – Will Sawin
    Commented Mar 20 at 18:35

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