# Diophantine equation without elementary solution but with simple non elementary solution

Is there some example of an diophantine equation that satisfies:

1. No solution is known using elementary methods.
2. It is simple to solve using non elementary methods (e.g. using number fields).

My goal is to find good motivation to dive into advanced algebra for someone who is used to solve everything using elementary methods, to show that something that is impossible to solve elementary is really easy using advanced techniques. Ideally if the person can try to attack the equation by himself, give up and then recognize the "simple" solution using advanced techniques and understand it (at least the main idea).

It is not a problem to find some equations as such in Number theory textbooks, but usually those are also solvable using elementary methods. And if there is an equation in which I am confident person will not solve it using elementary methods, it is something with quite complicated proof (extreme example would be Fermat's Last Theorem).

Update: For clarity, let's consider elementary to refer to methods known to Euler (or mathematicians at that time generally). As for simple solution using advanced techniques, that is definitely subjective, and I have currently no idea how to define this, but I believe there is some kind of consensus among mathematicians on things that are simple and elegant.

• I think some of Mordell curves are good examples. $y^2=x^3+k$ for $k=-1, -4, 2$ for example. – Ghartal Jun 14 '17 at 20:00
• No this one, I suppose? Even with elliptic curves it is non-trivial, but the actual solution (or its size) shows that elementary attempts are doomed to fail ... – Hagen von Eitzen Jun 14 '17 at 20:00
• Actually this is my question too. I am tired of elementary methods! – Ghartal Jun 14 '17 at 20:06
• @Ghartal Those look good, although I am not familiar with the solution and how complicated it actually is. – Sil Jun 14 '17 at 20:07
• @HagenvonEitzen That looks good, though if the solution is complicated, that won't do. I am thinking if what I ask is even possible, perhaps only problems with complicated proofs are those that do not have elementary solution... – Sil Jun 14 '17 at 20:08

First you should define "elementary" and "simple". Granted this is done, and we agree informally on these preliminaries, I think that a good example (although not "ridiculously" easy to solve using advanced techniques) would be the determination of the primes $p$ which are of the form $x^2 + ny^2$, for a given positive integer $n$. See the book by David Cox bearing the same title. Particular solutions were given by Fermat, Euler, Lagrange, Legendre, Gauss (can their methods be called elementary ? If not, consider only Fermat and his method of "descent"). The complete solution reads: "Let $n$ be a positive integer. Then there is an irreducible polynomial $f_n (X) \in \mathbf Z[X]$ such that for a prime $p$ dividing neither $n$ nor disc $f_n$, $p$ is of the desired form iff ($-n/p$)= $1$ and $f_n(X) \equiv 0$ mod $p$ has an integer solution".

Cox notes that although "the statement of the theorem is elementary, the polynomial $f_n$ is quite sophisticated: it is the minimal polynomial of a primitive element of the Hilbert class field of $K=\mathbf Q (\sqrt{-n})$. More precisely, the proof needs class field theory and higher reciprocity laws, and modular functions and complex multiplication (= explicit CFT over an imaginary quadratic field) are used to provide an algorithm to give an effective answer.

• The Cox's book is actually on my "to read" list (already tried but never got to the end...), and I agree the final statement looks "elementary". It is definitely nice problem to demonstrate advanced number theory techniques, but it does not satisfy the condition to be "simply" solvable by them (the whole book is practically preparing reader for that final solution...). As for the definition of elementary and simple, that is always a discussion, but I would say that for example anything solved by methods known in Euler's times can be safely considered elementary, perhaps... – Sil Jun 15 '17 at 17:49

I think a good example is given by the equation $$x^2-6y^2=1$$ This equation has infinitely many integer solutions $(x,y)=(a_n,b_n)$ determined by $$a_n+b_n\sqrt6=(5+2\sqrt6)^n\space\space........ (n\ge1)$$ Here the calculation of the number $5+2\sqrt6$ is obviously "fundamental" and it is not obvious at all that there are infinitely many solutions if this equation is looked from "elementary".

In general, for any square-free positive integer $d$, the equation (Pell's) $$x^2-y^2\sqrt d=1$$ also has infinitely many integer solutions given by $$a_n+b_n\sqrt d=(a_0+b_0\sqrt d)^n$$ where $a_0+b_0\sqrt d$ is called the fundamental unit of the quadratic field $\mathbb Q(\sqrt d)$

NOTE.- To take into account here that elementary is a relative concept in mathematics. For many people the text above is quite "elementary".

I accept informally your update on elementary vs. simple non elementary, but let us make it a bit more precise in order to go on. I guess everyone of us has at his disposal a favorite "tool kit" of concepts, theories, theorems, etc. which he uses to attack problems (solved or unsolved) and which depends on his formation. The point I want to stress is that every tool in such a kit is considered as already manufactured, ready for use without question, so that what you call a "simple non elementary" method just consists in a "ridiculously easy" (in your previous words) application of the tools contained in a kit labelled "non elementary". If you agree, then :

1) Concerning the diophantine equation $p = x^2+ny^2$, I think the solution explained in Cox's book belongs to the category "simple non elementary", the tool kit is just CFT, and all the more so when restricting to $n$ squarefree, $n$ not congruent to $3$ mod $4$ (this is just thm. 5.1 of the first chapter)

2) While we are at problems on sums of powers, here is a perhaps more convincing example. Sums of 2 squares belong to the elementary kit labelled "Fermat", sums of 4 squares to the (not so) elementary box called "Lagrange". And what about the Sylvester problem: when is a prime a sum of two rational cubes, $p=x^3 + y^3$ ? See e.g.

www.college-de-france.fr/site/don-zagier/lecon_inaugurale.htm

The answer is: (1) never for $p\equiv 2$ or $5$ mod $9$ (elementary kit "Fermat"); (2) always for $p\equiv4$ or $7$ mod $9$ (non elementary). In his course, Zagier gives 2 truly "gigantic" couples $(x, y)$ for $p= 382$ and $1789$ (the "french prime"), which tend to show that even a computer search for a negative answer would have been futile; (3) a necessary and sufficient (but not down to earth) criterion is available for $p\equiv 1$ mod $9$. The tool kit here could be labelled "Birch & Swinnerton-Dyer conjecture". More precisely, the Sylvester question is equivalent to whether the Mordell-Weil group $E_p (\mathbf Q)$ of the elliptic curve $E_p : x^3 + y^3 = p$ is non trivial, or else whether the rank $r_p$ of $E_p (\mathbf Q)$ is $>0$. According to BSD, this happens iff the $L$-function of $E_p$ vanishes at $s=1$. Although BSD is still a conjecture, enough partial results are known to show the properties (1) - (3) above. I stress that the BSD kit here contains tools related to the special value $L(E , 1)$ attached to an elliptic curve $E/\mathbf Q$, not to the specific Sylvester problem. Once the kit is granted, the case (2) is straightforward in your sense. But the case (3) is not simple, it is based on a theorem specific to the curve $E_p$ due to Villegas-Zagier.

Does this example cover all the aspects of your question about elementary/simple non elementary ?