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I have two questions.

$(1.)$ Is there a property of the natural numbers such that we know at least one number satisfies it but we don't know which one?

Even more,

$(2.)$ Is there a property of the natural numbers such that we know at least one number satisfies but we don't know which one and moreover we cannot bound any such number by another known number?

Now the question might arise of what does it mean to know a number. To be able to write it down in finite time given an infinite amount ink and paper would suffice.

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  • $\begingroup$ Well, I'm not sure if that will help, but I've heard that number theorists know that one of the numbers $3$, $5$ or $7$ (not sure now if these are the numbers, but there are three of them) is a primitive root $\operatorname{mod}\: p$ for an infinite amount of prime numbers $p$, but they don't know which one of them. $\endgroup$ – Shoutre Oct 24 '15 at 20:59
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    $\begingroup$ This question is similar but narrower, and has numerous good answers that fit well here too: math.stackexchange.com/questions/1315615/… $\endgroup$ – Travis Willse Oct 24 '15 at 21:19
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Goldbach's conjecture is the statement that, if $n$ is an even natural number and $n>2$ then $n$ is the sum of two primes. It is not known whether this conjecture is true, nor is any bound known on how big the first counterexample could be if the conjecture is false. In view of this situation, consider the following property of a natural number $n$: Either $n$ is the first counterexample to Goldbach's conjecture or Goldbach's conjecture is true and $n=0$. We know that there is exactly one $n$ with this property (because either Goldbach's conjecture is true or it has exactly one first counterexample), but we don't have any reasonable bound for how big $n$ could be. (The weasely word "reasonable" is there to exclude "bounds" like "$7+$ the first counterexample to Goldbach's conjecture or $13$ if Goldbach's conjecture is true".)

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  • $\begingroup$ I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $n\in \mathbb{N}$ we can answer whether $p(n)$ or $\neg p(n)$? If you can think of one feel free to share it :) $\endgroup$ – Anguepa Apr 16 '16 at 11:19
  • $\begingroup$ @Anguepa By "we can answer whether $p(n)$ or $\neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $\neg p(n)$? $\endgroup$ – Andreas Blass Apr 16 '16 at 19:00
  • $\begingroup$ Yes. Sorry I should've made that clear. $\endgroup$ – Anguepa Apr 17 '16 at 18:02
  • $\begingroup$ @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.) $\endgroup$ – Andreas Blass Apr 17 '16 at 23:02
  • $\begingroup$ This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed. $\endgroup$ – Anguepa Apr 18 '16 at 10:38
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For the first question, we know there is a smallest natural number $n$ such that $\pi(n)>\text{li}(n)$ where $\pi(x)$ is the prime counting function and li$(x)$ is the logarithmic integral, but we don't know what the number is.

EDIT: For the sake of having a more complete answer, I will include our discussion in the comments.

There has been much progress in determining an upper bound for the first integer at which this switch occurs, but still the number eludes us. Strangely enough, we know that there are infinitely many numbers for which this inequality holds.

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    $\begingroup$ This is what I had in mind; not only do we know there is a smallest, we know that $\pi(x)-\mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive. $\endgroup$ – Clayton Oct 24 '15 at 21:02
  • $\begingroup$ Very interesting @AJStas. Doesn't this answer the second question too? $\endgroup$ – Anguepa Oct 24 '15 at 21:06
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    $\begingroup$ @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time. $\endgroup$ – Clayton Oct 24 '15 at 21:08
  • $\begingroup$ @Anguepa There have been many upper bounds for this number though, so I wasn't really sure. $\endgroup$ – Rocket Man Oct 24 '15 at 21:08
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    $\begingroup$ @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$. $\endgroup$ – Clayton Oct 24 '15 at 21:12
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Here's an interesting example that is the subject of recent (and celebrated) research:

Consider the set $\{p_1, p_2, p_3, \ldots\}$ of prime numbers, listed in increasing order. The $n$th prime gap is the difference $p_{n + 1} - p_n$. In 2013, Yitang Zhang showed that there are infinitely many prime gaps of size $\leq 7 \cdot 10^7$. Since there are only finitely many positive integers smaller than this size, at least one of them must occur infinitely often, but we don't know which. Since then, the Polymath project has improved that upper bound to $246$ (as of last December). The famous and longstanding Twin Prime Conjecture says that $2$ occurs infinitely often.

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It is known that one of ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational, but no one knows which one:

W. Zudilin (2001). "One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational". Russ. Math. Surv. 56 (4): 774–776. doi:10.1070/RM2001v056n04ABEH000427.

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    $\begingroup$ Probably all of them are irrational. $\endgroup$ – azimut Oct 30 '15 at 21:20
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    $\begingroup$ Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have. $\endgroup$ – marty cohen Oct 30 '15 at 21:38
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Consider $\Sigma(m,n)$ to be the busy beaver function with $m$ states and $n$ colors.

The following is then very hard to find:

What is the smallest $m$ such that $\Sigma(m,2)>\Sigma(10,3)$?

Since there is a color reduction method for reducing a $k$ state, 3 color machine to a $7k$ state, 2 color machine, it is known that $m\leq 70$. On the other hand, it is clear that $m>10$. But it is very hard to say anything between this.

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Since we do not know , for example $\Sigma(5)$ , we do not know which number $n$ satisfies $n=\Sigma(5)$, but we know that some number $n$ must satisfy the property. In this case we do not even have an upper bound.

Another less trivial example : The $20$-th Fermat-number $2^{2^{20}}+1$ is known to be composite, but no factor is known.

Denote $n$ to be the smallest prime factor of the $20$-th Fermat-number. This number is well-defined but we do not know it, but in this case, we can bound it.

If this does not hit your intention, please precise which kind of properties you mean.

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  • $\begingroup$ what means $\Sigma$ $\endgroup$ – miracle173 Jun 10 '17 at 6:28

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