Here are a just a few examples where patterns which appear to hold, start to fail for large numbers. Of course, as humans, we have a biased view of “large” in the context of numbers.
- Consider the sequence, 12, 121, 1211, 12111, 121111, … etc. Are they all composite? Well the first 100 are. Is that enough evidence to be confident that they are all composite? Nope—you can find the first prime number in the sequence yourself, or see this post by Alon Amit, with this and other much better examples.
- Primes at most x are more often congruent to 2 mod 3 (as opposed to 1 mod 3) until x=608, 981, 813, 029. (I.e., more than 600 billion. See page 203 of “A Friendly Introduction to Number Theory” by Silverman, 4th edition).
- Fermat numbers, of the form, 2^(2^i) + 1 were known (by Fermat) to be prime for i=0, 1, 2, 3 and 4. So that’s only 5 numbers. But the fifth one, 65 537 is so big! Perhaps this influenced Fermat’s unfortunate (or perhaps fortunate?) decision to conjecture that they are all prime. Perhaps he even had a guess regarding the significance Gauss would find for his primes 150 or so years later. Well the next (at least) 28 Fermat numbers aren’t prime. Maybe there’s none others? (How embarrassing would that be?) Here is “compelling evidence” by Boklan and Conway that the ONLY Fermat numbers that are actually prime may be the first 5 Fermat numbers. Of course, since the paper of doesn’t prove that there are no more Fermat primes, it’s inclusion here is perhaps ironic. So it is currently conceivable that Fermat’s conjecture was not merely incorrect, but spectacularly so… Naturally it was the great Euler who originally factored the first composite Fermat number; 2^32+1=641×6700417.
- However, even the great Euler had his own, almost maximally incorrect conjecture. Indeed, he hypothesized that there are mutually orthogonal Latin squares of order n if and only if n is not 2 mod 4. It turns out that there are none for n= 2 (this is trivial), 6 (this is hard, and surely where Euler put in some effort); but there is one for all other values of n. (Together, Bose, Parker and Shrikhande showed there are mutually orthogonal Latin squares for all even n congruent to 2 mod 4, n>6). (See p. 122 “Introduction to Combinatorial Designs”, Wallis, 2nd edition).
- Graph labeling problems are a great place to find other potential “maximally incorrect” conjectures. Gallian has a wonderful survey of recent work, progressing so fast that it would otherwise be quite difficult to keep track of it.
Here’s one of personal significance: Conjecture 1 from this beautiful paper of Gray and MacDougall is very reminiscent of Euler’s conjecture.
The main part of this conjecture is that certain graph labelings exist; but the negative part–that some don’t–fail, completely, as soon as they stopped checking! Here’s the constructive proof. (I’m still proud of this contribution, with Jeremy Holden and James McQuillan.)
- Another example (this one famous) involves the product of sums of squares with a strikingly wrong conjecture. After Degen provided a formula for showing that the sum of 8 squares multiplied by the sum of 8 squares is again the sum of 8 squares, he believed, quite incorrectly, that there were similar larger formulas for sums of 2^k squares for any k.
Hurwitz proved that the (bilinear) formulas for sums of 2, 4, and 8 squares known to Degen were the only ones. Oops. (The analogous formulas for sums of 2 squares and sums of 4 squares are also significant for complex numbers and quaternions, respectively
- The circle division problem is often used as an excellent example of how small examples may suggest a striking, yet incorrect answer.
Given n points on a circle, a chord for each pair; chords divide the circle into how many regions? (The maximum number possible). For n=2, 3, 4, 5, 6 it’s 2, 4, 8, 16 and— 31. That’s 31 regions, not 32. The answer is the sum of binomial coefficients (n,4) + (n,2) + 1. It’s so tempting to think it’s a power of 2.
Here is an excellent video for this example by Grant Sanderson.
8. So now we wonder about our research:
The crossing number of the complete graph with n vertices is conjectured to be the square of a particular binomial coefficient for n odd. This seems very compelling, partly just because a binomial coefficient is a really simple expression for such a complicated question.
More precisely, if n=2m+1, crossing # of K_n is conjectured to be the square of the binomial (m,2). It’s true for many drawings, e.g., these ones.
And yet, the smallest unknown case is just m=6. With Pan and Richter, we provide a few clues for the first unknown case. My best shot at understanding so far, (with Richter, and work that I’m very proud of) is summarized here.
I expect this conjecture will be verified for m=6, but (personally) feel very unsure of the conjecture beyond m=6.
So in conclusion…
I hope you get the idea. When mathematicians seem obsessed with rigor, it is at least in part due to our history of making mistakes, seemingly every time we tried to jump to a conclusion. But perhaps there’s something more. So for that, we end with a quote from William Thurston:
what we are doing is finding ways for people to understand and think about mathematics… what they really want is usually not some collection of of ‘answers’-what they want is understanding.