1
$\begingroup$

Solvability of a general diophantine equation has been proved undecidable. As a famous example of knowledge, we know that $x^n+y^n=z^n$ has no solutions (in $\mathbb{N}$) for $n>2$. As a famous example of ignorance, we don't know whether or not $y=x+2$ has infinite prime numbers solutions.

Do you know some example of a simple Diophantine equation of which we ignore whether or not it has any solution?

In particular is there some "famous" two variable Diophantine equation of unknown solvability? Maybe restricting possible solutions to prime, or to some subset of $\mathbb{N}$

*Basing on wikipedia $$2^{c-1}\equiv1\,(\mathrm{mod}\,c^2)$$ is not known to have composite solutions. This is the best I found so far.

$\endgroup$
11
  • 1
    $\begingroup$ What is unknown about solutions in $\mathbb N$ of $x^n+y^n=z^n$? $\endgroup$ Jun 2, 2019 at 13:14
  • 2
    $\begingroup$ I am not sure what the exact question is... $\endgroup$ Jun 2, 2019 at 13:15
  • 2
    $\begingroup$ There are very many such equations. There are no general methods for solving all equations over the integers--only special interesting families, like elliptic curves. $\endgroup$
    – Dzoooks
    Jun 2, 2019 at 13:24
  • 1
    $\begingroup$ An adjustment of yours: Does $2^x-1=y$ have infinitely many solutions for prime $x$ and $y$. It is easily verified that $y$ is prime implies $x$ is prime. The converse is false: $2^{11}-1=2047=23\cdot 89$. But we may still ask for infinitely many prime $x$ and $y$. These are the Mersenne Primes. $\endgroup$
    – Dzoooks
    Jun 2, 2019 at 13:35
  • 2
    $\begingroup$ It is not known whether there exist integers $A,B,C$ with $A>0$ such that $\{Ax^2+Bx+C:x\in \Bbb N\}$ contains infinitely many primes $\endgroup$ Jun 2, 2019 at 19:43

2 Answers 2

2
$\begingroup$

If you're asking for solutions where $x$ and $y$ is prime, there are none. $x^2$ diving $2^x$ immediately fails - note that if $x$ is prime, we have $2^x \equiv 2 \pmod x$ by Fermat's little theorem, and so $2^x - 1 \equiv 1 \neq 0$.

The equation you've given, $x^n + y^n = z^n$, has been solved for all naturals $n$ - there are trivial solutions for $n=1$, Pythagorean triads have been fully parametrised, and no solutions for $n > 2$ exist due to Fermat's Last Theorem, which was proven by Andrew Wiles in 1994.

An example of a simple statement which is still unsolved is whether infinitely many integers $n$ exist such that $n \mid 2^n + 3$. A more famous problem, although not Diophantine, is whether there exist infinitely many twin primes.

EDIT: the OP has since changed the equation from $2^x - 1 = yx^2$ to $2^x - 1 = yz^2$. In this case, it is unlikely that we can tell if there are infinitely many solutions - if we relax the prime condition on $z$ to allow $z=1$ the problem simply becomes finding Mersenne primes, for which there exists no simple algorithm to find as of current mathematical/computational techniques.

$\endgroup$
3
  • $\begingroup$ Ok I complicated my example adding a variable. Your example regard counting the solutions, but I am searching for an equation not involving anything than arithmetic operators, which we still don't know if it has any solution. $\endgroup$
    – J.Ask
    Jun 2, 2019 at 13:35
  • $\begingroup$ @J.Ask Fermat's Last Theorem states that there are no such solutions for $n>2$. I don't know what you mean by only arithmetic operators or how that contradicts what the theorem states. $\endgroup$
    – auscrypt
    Jun 2, 2019 at 13:38
  • $\begingroup$ Fermat's Last Theorem is an example of knowing that there are no solutions. Many example can be found in which we don't know if the solutions are infinite. I'm searching for an example of not knowing if there is any solution, about an equation involving only the four arithmetic operations and natural powers and natural roots $\endgroup$
    – J.Ask
    Jun 2, 2019 at 13:53
1
$\begingroup$

One of the simplest-to-state unsolved problems in mathematics is the Collatz conjecture:

Consider the algorithm:

  1. Choose a positive integer to start.
  2. If the integer is even, divide it by $2$.
  3. If the integer is odd, multiply by $3$ and add $1$.
  4. Repeat.

For every integer tried, following this algorithm eventually leads you to the cycle $4 \to 2 \to 1 \to 4\ ...$ The Collatz conjecture is that every starting integer will lead to this cycle. There might be some other repeating cycle out there, but none has ever been found.

$\endgroup$
7
  • $\begingroup$ I should point out, though, that the Collatz conjecture is not considered a important result. Presumably a solution may be adaptable to similar algorithms, but it isn't likely to have much of an impact. Other conjectures, particularly the Riemann Hypothesis, are of far more importance. But they are also much more complex to state. $\endgroup$ Jun 3, 2019 at 1:23
  • $\begingroup$ it would answer if $x^n$ can get you to x via cubing and multiplying by x, and sqrting. $$6=\sqrt{\sqrt{\sqrt{\sqrt{6{\sqrt{6^{10}}}^3}}}}$$ for example. $\endgroup$
    – user645636
    Jun 3, 2019 at 1:56
  • $\begingroup$ @PaulSinclair Why do you say that? Just because you've heard it from someone? I don't think anyone has the slightest idea what impact any solution to any problem will have in the future. It's pity to hear such statements, because it might undermine the importance of a solution to such a problem. And just because more people have tried to solve the Riemann instead of the Collatz doesn't mean it is any more important. $\endgroup$
    – user366820
    Jun 3, 2019 at 2:42
  • $\begingroup$ @NaturalNumberGuy - I've heard estimates that as much of half of all published number theory results in the last century or so are dependent on the Riemann hypothesis. Discovering it is false would fundamentally change our understanding of the natural numbers. That is why it is so important. On the other hand, the Collatz conjecture is just a funny thing someone threw together that no one has been able to solve. There are no significant results riding on it. The great conjectures are those who fundamentally shape our understanding of some field. The Collatz conjecture is not amonst them. $\endgroup$ Jun 3, 2019 at 15:39
  • $\begingroup$ @PaulSinclair What if the Collatz Conjecture in it's heart is related to a proof of the Riemann hypothesis? Ever thought about that? I do not fully agree with your argument, but maybe you are like those who jump on the global climate change wagon for no reason. Knowing how many primes there are up to a given quantity sure is interesting, but a solution to something we do not know why it's there I think is even more interesting. $\endgroup$
    – user366820
    Jun 4, 2019 at 2:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.