I was recently reading up about computational power and its uses in maths particularly to find counterexamples to conjectures. I was wondering are there any current mathematical problems which we are unable to solve due to our lack of computational power or inaccessibility to it.

What exactly am I looking for?

Problems of which we know that they can be solved with a finite (but very long) computation?

(e. g. NOT the Riemann hypothesis or twin prime conjecture)

I am looking for specific examples.

  • 2
    What exactly are you looking for? Problems of which we know that they can be solved with a finite (but very long) computation? (e. g. not the Riemann hypothesis or twin prime conjecture) – Carlos Esparza Aug 16 '19 at 20:49
  • @stackupphysics I think you need to clarify whether "lack" refers to a technological insufficiency (e.g. we don't yet have enough processing power) or a theoretical insufficiency (e.g. even a perfect computer could never solve the problem). – Jam Aug 16 '19 at 21:22
  • @0x539 I'll update –  Aug 17 '19 at 04:18
  • 7
    Near duplicate on MO https://mathoverflow.net/q/112097/30186 – Wojowu Aug 17 '19 at 09:39
  • 4
    Perhaps Skewes' number, the least $x$ for which $\pi(x)\le li(x)$. – DanielWainfleet Aug 17 '19 at 11:47
  • 3
    The first thing I thought of is chess — 8-piece tablebases are currently on the bubble, and position counts at ply N are only up to N=13 or something — but not sure that’s “math” by your definition. – Jeff Y Aug 17 '19 at 12:42
  • @JeffY seems interesting can you elaborate more. Maybe as an answer it would be quite a treat to read up –  Aug 17 '19 at 15:24
  • 2
    What is the largest prime number with 500 digits? – Asaf Karagila Aug 17 '19 at 18:22
  • Not exactly what the OP is looking for, but OEIS is full of sequences where less than ten terms are known and the next unknown term requires too much time/memory to compute with current technology – Yuriy S Aug 17 '19 at 18:38
  • 1
    With proof assistants and a lot of computational power, you can explore all proofs (roughly sorted by size). As long as you manage to state a conjecture in terms the assistant understands, you can just wait until it finds a proof for it (you might prove the twin prime conjecture while waiting for Riemann, although you won't notice). Were you specifically looking for examples where we *almost* have enough computational power? – Marc Glisse Aug 17 '19 at 19:24
  • @MarcGlisse I am looking for things which can be solved with current understanding like no special proof assistant need to be programmed however we are unable to so due to lack of computation cover –  Aug 18 '19 at 05:08
  • @Jeff Y it seems this question is rather trivial if something like this works. What is the most efficient optimal chess strategy? – marshal craft Aug 18 '19 at 08:15
  • Or prime digits as Asaf Aragila said. End of discussion. Or, where a position of a particle bouncing around a box in such amount of time. – marshal craft Aug 18 '19 at 08:19
  • Really the result or output of any algorithm after such time given such memory. – marshal craft Aug 18 '19 at 08:22
  • 3
    As @MarcGlisse mentions, there is a simple algorithm which simply lists all statements provable in ZFC, so with sufficient computational power you could prove anything that's provable. However, something like the Riemann hypothesis can't necessarily be resolved this way, since it might be independent from ZFC, in which case neither it nor its negation will ever be proven by the program. Indeed, the class of unsolved problems the OP is interested in is precisely the class of open problems which are known to not be independent from ZFC. – Jim Belk Aug 18 '19 at 09:12
  • @JimBelk I don't know too much about ZFC but what you say is quite what I'm looking for. I'll update my question with these additional details –  Aug 18 '19 at 13:24
  • @YuriyS can you share a link where I can read more about this. I am not exactly sure if it's a suitable answer as I haven't studied about this but I would definitely love to read about it if you did write an answer –  Aug 18 '19 at 13:32
  • 1
    @AsafKaragila http://m.wolframalpha.com/input/?i=largest+prime+below+10%5E500 :) – Wojowu Aug 18 '19 at 13:37
  • What is ZFC? [extra chars] – spacetyper Aug 19 '19 at 04:34
  • @spacetyper I'll add a few links for you to read up –  Aug 19 '19 at 05:12
  • There's loads of stuff we can't bait nature into computing even when you use limiting physical constants of what we can do (which we are many orders of magnitude from hitting), "even with all the energy in the universe (it doesn't matter what you use, mass-energy is often used) you'd die before the calculation finished" -remember this is what computation really is, tricking nature into working stuff out. The other thing is what you call "solved" - typically we /can/ enumerate states surjectively (think Z^2 onto Q by (a,b)->a/b modulo zeros blah blah) easily, brute force from there. – Alec Teal Aug 19 '19 at 10:28
  • I should add that "brute force" is generally crap, consider sorting by asking "is it sorted?" (for each card from the 2nd forward check the previous card < this card) - if not, permute two cards, I'm going to cop out here by saying "permute randomly". It'll still get there eventually. – Alec Teal Aug 19 '19 at 10:30
  • 2
    @Asaf if that's not known it's because it's irrelevant not because we lack computational power. :-) Testing numbers with a couple hundred decimal digits is routine (how would you want to do 4096 bit RSA otherwise?) and just start form the largest number eligible and iterate. Prime Number Theorem predicts that you won't have to test too many (roughly three times the number of digits). [Yes, yes, you point stands but I am a nit-picker.] – quid Aug 19 '19 at 11:00
  • [Is there any conjecture that we know is provable/disprovable but we haven't found a proof of yet?](https://math.stackexchange.com/q/2273525/26369) is very closely related, though only has a couple specific examples – Mark S. Aug 19 '19 at 11:41
  • 1
    @quid: Well, it's my conjecture (and arguably a very important one) that the least significant digit of the largest prime with at most $10^{100^{1000^{10000}}}$ digits is $3$, and I expect this to be proved soon. – Asaf Karagila Aug 19 '19 at 13:33
  • 1
    An interesting related result: A Relatively Small Turing Machine Whose Behavior Is Independent of Set Theory - https://arxiv.org/abs/1605.04343 – Harald Thomson Aug 20 '19 at 06:09
  • 1
    This is actually fairly common in cryptography. Some block ciphers use an S-box, which is a public permutation with a small size (e.g. 8 bits). We don't have enough computing power to find an ideal 8-bit S-box as it would require searching through $256!$ different S-boxes. With a 4-bit S-box, it's only $16!$ which is [possible to analyze](https://eprint.iacr.org/2011/218.pdf). – forest Aug 20 '19 at 06:27
  • For those interested in Proof Assistants, there is a new proposed SE site [ProofAssistants](https://area51.stackexchange.com/proposals/126242/proof-assistants?referrer=Njg4YTJmMjYwOTIxNjdkNGEyMmZkNzE0Y2M4YmFhOTY3OWVmNDUwNWM3ZmFlMjYwYTRiYzZiZWY1ODg5ZjdiMqfBIyhShuHO9QbGuJfVwOYRvfVgPfbJQrn2UOFTeOe-0) – Guy Coder Nov 28 '21 at 10:31

18 Answers18


Goldbach's weak conjecture isn't a conjecture anymore, but before it was proved (in 2013), it had already been proved that it was true for every $n>e^{e^{16\,038}}$. It was not computationally possible to test it for all numbers $n\leqslant e^{e^{16\,038}}$ though.

José Carlos Santos
  • 397,636
  • 215
  • 245
  • 423

Some notorious problems of this kind are in discrete mathematics but involve a search space that is many magnitudes beyond what is feasible. For example, the values of certain Ramsey numbers or the existence of a Moore graph of degree 57.

  • 1,434
  • 1
  • 10
  • 15
  • 15,230
  • 1
  • 18
  • 36
  • 33
    Ramsey numbers are the canonical “unfeasible computation”. As Erdős said, [paraphrased by Joel H. Spencer](https://en.wikipedia.org/wiki/Ramsey%27s_theorem#cite_note-7): “Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5, 5) or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for R(6, 6). In that case, he believes, we should attempt to destroy the aliens.” On the other hand [cont’d], – Peter LeFanu Lumsdaine Aug 17 '19 at 10:08
  • 4
    [cont’d] …values of Ramsey numbers may not quite be what the OP is after, since they’re not typically considered as “counterexamples to a conjecture” (although of course they can be seen as such, in contrived ways). – Peter LeFanu Lumsdaine Aug 17 '19 at 10:09
  • 4
    @PeterLeFanuLumsdaine: That is shockingly delightful! – Daniel R. Collins Aug 18 '19 at 04:24

Historically, a very important, computationally intensive problem arising from physics was lattice QCD (LQCD). LQCD is a theoretical framework for computing basic quantities like the mass of the proton, and it was introduced by Ken Wilson back in the 70's. However, after some initial successes, this approach stagnated due to a lack of computer power. The basic problem is that our universe has an obnoxiously large number of dimensions (four, in case you were wondering), and doing integrals in four dimensions takes an insane amount of memory. I heard a story that Ken Wilson gave a talk at a conference on LQCD where he declared that "Lattice QCD is dead" as long as a certain 4D integral could not be computed, as was the case at the time he said this.

Several years (or decades) later, computer technology matured to the point that said integral could be computed, and then LQCD theory picked right back up where it left off. Today it is again a flourishing discipline. However, other problems arising from LQCD continue to push supercomputer technology. Apparently LQCD is used as a benchmark for supercomputers nowadays.

  • 14,399
  • 4
  • 29
  • 70

If you are including games as part of “math”, chess provides some nice unsolved problems due to computational limits. The game of chess itself cannot even be weakly solved (https://en.m.wikipedia.org/wiki/Solved_game#Overview). But strong solutions are known for a subset of chess positions, those with seven or fewer (total) pieces on the board. These are called (endgame) tablebases: https://en.m.wikipedia.org/wiki/Endgame_tablebase#Background. Any position with eight or more pieces is currently at or beyond present computational resources (chess games start with 32 pieces).

Another source of difficult computation around chess is counting total positions (of certain types) after a certain number of moves. Such as the number of chess games ending in checkmate in exactly N plies (moves by one side), which is presently only known for N <= 13: https://oeis.org/A079485. Or just the total number of possible chess games consisting of N plies, which is presently only known for N <= 14: https://oeis.org/A048987.

Jeff Y
  • 113
  • 1
  • 7
  • 4
    Completely unimportant, but the number of games is known up to 15 ply: http://wismuth.com/chess/statistics-games.html – A. Rex Aug 18 '19 at 15:55
  • 3
    What is your source for "the game of chess itself cannot even be weakly solved"? The article you linked itself calls this "speculation", which it might well be despite the sprawling game tree. – kviiri Aug 19 '19 at 17:43
  • 1
    @kviiri Probably it means "cannot currently be weakly solved", i.e. no one has done it yet or currently has a good plan to do it. – Will Sawin Aug 20 '19 at 03:25
  • 1
    @kviiri Yes, the context is "with today's computational (and storage) resources". As the article notes, any finite game is **theoretically** solvable. But considering the known effort and resources it took to weakly solve checkers, and knowing that chess is an orders-of-magnitude larger game... Nowhere close. – Jeff Y Aug 22 '19 at 13:34
  • 1
    Here's another explanation of why any claim of (weakly) solving even most chess openings (subsets of chess) in the current day is **always** firmly in the realm of spoofs: https://en.chessbase.com/post/the-chebase-april-fools-revisited They had published the spoof, which fooled many people, including me, a few days earlier: https://en.chessbase.com/post/rajlich-busting-the-king-s-gambit-this-time-for-sure – Jeff Y Aug 22 '19 at 13:52

Packing problems come to mind, i.e. how to achieve the densest packing of some kind of geometric objects, such as spheres or dodecahedrons. The interesting thing is that this is not a discrete problem, as there are uncountably many irregular, non-periodic packings that need to be checked. Still, the original proof of the sphere packing problem managed to turn this into a finite number of linear programming problems which then could be solved on a computer.

In theory you can use the same approach for objects other than spheres or in higher dimensions (and indeed people do), but in practice you reach a point quite soon, where there is simply not enough computing power to solve the resulting problems.

  • 3,841
  • 17
  • 21

Is $e^{e^{e^{79}}}$ an integer? See this question for some background. Many other problems of this type are also technically unsolved, although the answer is almost definitely "no". This can be verified by a finite computation, but the sheer size of the numbers involved means that this is not feasible at the moment.

Note: as pointed out by @ruakh, if $e^{e^{e^{79}}}$ were, in fact, an integer, then a naïve finite computation would not be able to resolve the question. [Of course, this seems highly unlikely, but it is not known to be false absent proof.]

  • 16,219
  • 4
  • 25
  • 58
  • Nice to know about this. Thanks for pointing it out –  Aug 18 '19 at 13:29
  • 2
    This answer is thought-provoking (+1), but as far as I can tell it's not actually correct: if the answer is "no" then that could be verified by a finite computation that approximated it closely enough to rule out the neighboring integers; but if the answer is "yes" then there's no obvious computational way to prove it. All we could show is that it's within *ε* of an integer for ridiculously small values of *ε*. (In fact, this very issue was the motivation for the question you link to: see [https://math.stackexchange.com/a/13052/19345](https://math.stackexchange.com/a/13052/19345).) – ruakh Aug 20 '19 at 05:53
  • @ruakh Wow, I hadn't thought of that. You're right, I will edit the answer to reflect that later. – YiFan Aug 20 '19 at 10:23

It is strongly believed that the second Hardy-Littlewood conjecture is false, because it contradicts the first Hardy-Littlewood conjecture, which has the backing of not only the probabilistic heuristic but also a lot of recent work. The second link even states that if the first conjecture (also called the prime $k$-tuples conjecture) holds, then there are in fact infinitely many positive integers $x$ such that $π(x+3159)-π(x) = 447 > 446 = π(3159)$. This is obviously something that can be verified with sufficient computational power (simply test every positive integer $x$ until you find one that satisfies the desired inequality), but clearly it has not been done yet otherwise we would have heard news of it!

  • 53,491
  • 7
  • 84
  • 231
  • Interesting to get to know about this. Haven't heard about this conjecture before –  Aug 18 '19 at 13:28
  • I don't think this answers the question. It's structurally very similar to disproving the Riemann hypothesis, which is explicitly given as an example of something the OP doesn't want. – Brady Gilg Aug 19 '19 at 16:26
  • @BradyGilg: Note that I answered the question before it was significantly changed. Besides, this is quite different from disproving RH because people believe RH is true (so it shouldn't be disprovable) whereas people believe that the conjecture stated in my post is true and hence provable by a finite computation. – user21820 Aug 19 '19 at 17:00

Euler's conjecture that it takes $n$ $n$th powers to sum to an $n$ power is true for $n=3$ but proven false for $n=4,5$, for example,

$$27^5+ 84^5+110^5+ 133^5= 144^5\qquad\text{(found in 1966)}$$ $$95800^4 + 217519^4 + 414560^4 = 422481^4\qquad\text{(found in 1988)}$$

but nobody knows if it is false for any or all $n\geq6$. There are heuristics that suggest,

$$x_1^6+x_2^6+\dots+x_5^6 = z^6$$

has positive solutions as well and a fast enough computer might find it. For the moment, such computational power is not available to individuals.

Tito Piezas III
  • 47,981
  • 5
  • 96
  • 237
  • 5
    Interesting problem, but it does not seem to match: "Problems of which we **know** that they can be solved with a finite (but very long) computation? " (my emphasis). – quid Aug 18 '19 at 11:32

Optimal sorting networks for $n>10$.

For small, fixed numbers of inputs n, optimal sorting networks can be constructed, with either minimal depth (for maximally parallel execution) or minimal size (number of comparators)... The following table summarizes the known optimality results:

$$ \begin{array}{l|ccccccccccccccccc|} \hline n & 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14& 15& 16& 17 \\ \hline \text{Depth} & 0& 1& 3& 3& 5& 5& 6& 6& 7& 7& 8& 8& 9& 9& 9& 9& 10 \\ \hline \text{Size, upper bound} & 0& 1& 3& 5& 9& 12& 16& 19& 25& 29& 35& 39& 45& 51& 56& 60& 71 \\ \hline \text{Size, lower bound (if different)} & & & & & & & & & & & 33& 37& 41& 45& 49& 53& 58 \\ \hline \end{array} $$

  • 13,081
  • 1
  • 11
  • 35

The order of every finite projective plane is a prime power. If this is false, a counterexample can be constructed by exhaustive search of all non-prime powers in increasing order. This has been done by hand for $n=6$ and by computer for $n=10$, but as far as I know, $n=12$ is still out of reach, or at least, it hasn't been done.

  • 51,472
  • 6
  • 32
  • 63
  • 5
    My understanding of the question suggests that this isn't an example unless there's some upper bound at which point we can stop. My interpretation of the question is the problems need to be computationally decidable, not semidecidable. Otherwise, we could say of *any* (formal) theorem that "if it's 'true', then we can enumerate proofs until we find its proof". Of course, you could say the $n=12$ case *is* decidable, but it still seems to miss the spirit of the question unless there's something special about this case. I could just as well ask if a theorem has a proof of size at most $N$. – Derek Elkins left SE Aug 18 '19 at 01:57
  • An alternative conjecture used to be that a finite projective plane exists for every order not ruled out by the [Bruck-Ryser theorem](https://en.wikipedia.org/wiki/Bruck–Ryser–Chowla_theorem#Projective_planes). Orders 6 and 14 are so ruled out, but orders 10 and 12 are not. This conjecture is, of course, dead now that a plane of order 10 has been ruled out by computational means. But I think many people still would like to know whether a plane of order 12 exists. – Will Orrick Aug 18 '19 at 12:23
  • Is it believed that there is a finite projective plane with order not a prime power? – user21820 Aug 19 '19 at 15:03
  • @user21820 I'm not a worker in the field, so I can't really say. In 1995, Gary Mullen suggested it as a candidate for ["the next Fermat problem"](https://link.springer.com/article/10.1007%2FBF03024365) Unfortunately, Springer doesn't make the whole article available publicly. The MOLS problem he discusses is equivalent to the finite projective plane problem. See http://pi.math.cornell.edu/~web4520/CG9-0.pdf – saulspatz Aug 19 '19 at 15:18
  • Thanks! Upon accessing the full paper, it says that: Kirkman established the existence of projective planes of prime power order around 1850, and Bose showed the equivalence of having $n - 1$ MOLS of order $n$ and having a projective plane of order $n$. Bruck and Ryser proved that if $n$ is $1$ or $2$ mod $4$ and the square-free part of $n$ has at least one prime factor of the form $4k + 3$, then there cannot be $n-1$ MOLS of order $n$. This result was proven using numerous number-theoretic techniques, including Lagrange's 4-square theorem. – user21820 Aug 19 '19 at 15:48
  • However, the article does not say anything about whether it is believed that there are only finite projective planes of prime power order. It merely says: This problem has resisted all attempts at solution, including those using techniques from number theory, group theory, combinatorics, algebraic coding theory, and, more recently, computational approaches. And there are applications of the solution, regardless of whether the conjecture is found to be true or false. – user21820 Aug 19 '19 at 15:53
  • 1
    @user21820 Some remarks: (1) In one formulation the problem is to construct a $(q^2+q+1)\times(q^2+q+1)$ matrix $M$ with elements 0 and 1 having $(q+1)$ 1s in every row and column and satisfying $MM^T=qI+J$,where $J$ is the all 1s matrix. (2) If one requires only that $M$ have rational elements in the relation $MIM^T=qI+J$ then the Hasse-Minkowski criterion for the rational equivalence of quadratic forms given by $I$ and $qI+J$ implies that such an $M$ exists if and only if the Bruck-Ryser conditions are satisfied. Ryser attempted to strengthen the Bruck-Ryser result by incorporating... – Will Orrick Aug 27 '19 at 13:46
  • ... some of the additional structure that $M$ needs to have before applying the Hasse-Minkowski criterion, but this doesn't appear to lead to anything better. In short, no general existence criterion apart from the Bruck-Ryser conditions has ever been found, despite strenuous efforts. (3) Taking points to be lines through the origin and lines to be planes through the origin in $\mathbf{F}_q^3$, where $\mathbf{F}_q$ is a finite field, one obtains a projective plane of order $q$. Desargues theorem holds only for planes constructed in this way. (4) For $q$ prime, only the Desarguesian planes... – Will Orrick Aug 27 '19 at 13:47
  • ...are known, but for primes $q\ge11$ there is no proof that there aren't non-Desarguesian planes as well. For the prime powers 9 and 16, non-Desarguesian planes are known. One can adopt at least two different points of view: (A) if there is no general obstruction to the existence of a certain type of object, then examples of that type of object should exist; (B) if there is no general principle leading to the construction of a certain type of object, then it is unlikely that objects of that type exist. I don't know of any good argument for either point of view in this case. – Will Orrick Aug 27 '19 at 13:54
  • @WillOrrick: Thanks for the additional information! I'd personally lean towards (A) if there are infinitely many 'chances' and probabilistic heuristics favour existence. But I'd lean towards (B) in other cases. =) – user21820 Aug 27 '19 at 15:03

Littlewood proved in 1914 that there exists a number $n\in\mathbb{N}$ (called Skewes' number) such that:

$$ \pi(n) > \operatorname{li}(n), $$

where $\pi(n)$ is the amount of primes below $n$ and $\operatorname{li}(n)$ denotes the logarithmic integral $\displaystyle \int_0^n \frac{dt}{\ln t}$.

It is conjectured that $n$ is a huge number, recent analysis suggests $n\approx e^{727.951}$. Since then, researchers have worked to find lower and upper bounds for $n$. Currently it is held that:

$$ 10^{19}<n<e^{727.951}. $$

No such number has been found yet.

  • 36,226
  • 15
  • 79
  • 126
  • 4,788
  • 5
  • 29
  • 64
  • Is it actually believed that Skewes' number is $\approx e^{727}$? I thought I'd previously read that *a* crossing point is there but not necessarily the least crossing point (Skewes'), which as you point out could be as low as $10^{19}$ – Jam Aug 20 '19 at 00:31

There was the question: Are there m consecutive positive integers from k to k+m-1 which contain more primes than the m integers from 2 to m+1?

The problem itself is unsolved, but there is a hypothesis with the twin-prime hypothesis as the simplest special case:

Given n ≥ 2, and n integers $0 = k_1 < k_2 < ... < k_n$, and for every prime p ≤ n the set of remainders $k_i \mod p$ has fewer than p elements, then there are infinitely many integers p such that $p + k_i$ is prime for every 1 ≤ i ≤ n.

If there are n primes from 2 to 2+m-1, and we find $k_1$ to $k_{n+1}$ with $k_{n+1} ≤ m-1$, then the hypothesis is that there are infinitely many sequences of m consecutive integers containing n+1 primes.

Finding such a sequence was quite hard but was done. I think there are sequences known that point to 5 more primes in m consecutive integers than in 2 to m-1, but beyond that it's limited by processing power (or by willingness to use that processing power).

  • 8,131
  • 16
  • 33
  • 1
    The 1st paragraph is called the Hardy-Littlewood 2nd conjecture, the 3rd paragraph id called the Hardy-Littlewood 1st conjecture, see the answer from user21820. – Gerry Myerson Aug 20 '19 at 00:52

The number of distinct magic squares, for deceptively small sizes

A magic square of order $n$ is a square grid of $n \times n$ boxes where each box contains one distinct integer from the interval $[1 .. n^2]$, so that the sums of the numbers on each row, on each column and on each of the two diagonals are equal to each other. They have been studied for millenia by mathematicians in China, India and Persia, and continue to be of interest to both hobbyist and professional mathematicians.

The smallest magic squares, excluding the trivial case where $n = 1$, are of order $3$. This is one of them:

\begin{array}{|c|c|c|} \hline 8 & 3 & 4 \\ \hline 1 & 5 & 9 \\ \hline 6 & 7 & 2 \\ \hline \end{array}

In a sense, this is the only solution to the problem of this size: the other 7 magic squares of order 3 are mirrored and/or rotated versions of this grid.

We know the number of magic squares of orders 3, 4 and 5. The number of magic squares of order 6 is not known, but is believed to be in the order of $10^{19}$. The number of magic squares is not known for any order greater than 6 either. It should be noted that constructing magic squares of odd and doubly-even (divisible by four) orders is generally regarded as a simpler feat than constructing magic squares of singly-even orders like 6, although this may not guarantee the ease of enumerating all magic squares of such order over enumerating those of orders of smaller singly even numbers.

This problem is trivially solvable if the computational power constraint wouldn't stop us: we could just enumerate all $36!$ possible ways to fit the numbers in the grid, and check each for magic number property. In practice, we can apply a fair bit of pruning to explore only a small fraction of this space. We know the sum that should appear on each row/column/diagonal and we know that only an eighth of the configurations need to be checked to account for their mirrored and/or rotated copies; these and further insights or heuristics may be enough to make the problem computationally tractable for a well-supplied research effort in the coming years.

However, this is in a sense cop-out; even if we solve the number of magic squares of order 6, we'll still be left wondering what the number of magic squares of order 7 and greater might be --- that is, unless someone figures out a more efficient way to compute it than raw enumeration.

  • 1,220
  • 8
  • 12
  • Reading this Parker's Square comes to my mind if you know what I mean –  Aug 19 '19 at 18:44
  • The reference if you didn't get it- http://www.bradyharanblog.com/the-parker-square –  Aug 19 '19 at 19:41

I believe a problem connected with Graham's number is one of the things you are looking for. It is an upper bound to problem in Ramsey theory, that looks for a number $N$ satisfying certain criteria. I do not know much about that, but you can read more here https://en.wikipedia.org/wiki/Graham%27s_number .

But from my understanding, there are bounds on the number $N$, however the range of possible values derived from those bounds is still enormously large, way beyond computational possibilities of today (and probably ever). But with arbitrarily large, yet still finite computational power, the problem could be solved.

The lower bound is currently (as of 2021) only 13, leading a large gap between 13 and G to be improved by computation.

Alex Meiburg
  • 2,414
  • 9
  • 21
  • 1,007
  • 5
  • 12

What are the odds in Klondike Solitare? An attempt was made based on perfect knowledge yielding 79%, but the player doesn't have perfect knowledge. There's a bunch of Monte-Carlo results on that site; but a direct attack is far beyond reach, and it's not even known if the strategy they're using is actually optimal.

"One of the embarrassments of applied mathematics that we cannot determine the odds of winning the common game of solitaire."

  • 326
  • 3
  • 11

The Great Internet Mersenne Prime Search (GIMPS) comes to mind. It's all about using computational muscle to find larger Mersenne primes.

Mersenne primes are primes of the form $2^n-1$. There are $51$ which are known, including the largest known prime number. But there are conjectured to be infinitely many.


One example is the diophantine equation $x^y+y^z=z^x$ for positive integers $x, y, z$. In the paper https://www.researchgate.net/publication/267106572_On_the_Diophantine_equation_ax_y_by_z_cz_x_0 they proved that all the positive integer solutions satisfy $max(x,y,z) \leq e^{e^{e^5}}$. Thus the problem is reduced to a finite but very large computation.


Prime gaps - https://en.wikipedia.org/wiki/Prime_gap - of maximum known merit are found by increasing CPU (and GPU) computational power on the gapcoin network. See: https://gapcoin.club

... So the difficulty will simply be the length of the prime gap?

Not exactly. The average length of a prime gap with the starting prime p, is log(p), which means that the average prime gap size increases with lager primes. Then, instead of the pure length, we use the merit of the prime gap, which is the ratio of the gap's size to the average gap size.

Let p be the prime starting a prime gap, then m = gapsize/log(p) will be the merit of this prime gap.

Also a pseudo random number is calculated from p to provide finer difficulty adjustment.

Let rand(p) be a pseudo random function with 0 less than rand(p) less than 1. Then, for a prime gap starting at prime p with size s, the difficulty will be s/log(p) + 2/log(p) * rand(p), where 2/log(p) is the average distance between a gap of size s and s + 2 (the next greater gap) in the proximity of p.

When it actually comes to mining, there are two additional fields added to the Blockheader, named “shift” and “adder”. We will calculate the prime p as sha256(Blockheader) * 2^shift + adder. As an additional criterion the adder has to be smaller than 2^shift to avoid that the PoW could be reused. ...

Source: gapcoin.org



... (Twin prime conjecture) ...

... This is a just a special case of a far-reaching conjecture of Hardy and Lit- tlewood describing the frequency of prime gaps of any sizes. The Hardy- Littlewood conjecture predicts not only how often twin primes occur, but also how often any finite tuple of the form ( n + h 1 , n + h 2 , . . . , n + h k ) consists en- tirely of prime numbers. But even though analytic number theorists have be- lieved for many years that they know the answers to these questions, progress towards proving the existence of small gaps between primes has been slow. As recently as 2005 , the problem of establishing infinitely many bounded gaps be- tween primes was considered by many mathematicians to be “hopeless” ...



List of unsolved problems in mathematics: https://en.wikipedia.org/wiki/List_of_unsolved_problems_in_mathematics

  • The OP wants a problem that is known to be solvable with a finite algorithm, which afaik is not known to exist for finding either the primegap of maximum merit nor the twin prime conjecture. – Jam Aug 20 '19 at 00:35