# What makes a math problem, like Fermat's Last Theorem, "difficult"?

What makes the proof to Fermat's Last Theorem difficult? Because the proof uses vocabulary that is rich in information?

The theorem itself seems so intuitively correct, that one could almost believe it solvable with logical calculus! More generally: what makes a math problem difficult?

• That's a good question. A problem could be difficult because you haven't learned it yet. Adding fractions is generally considered difficult, probably by a majority of the population. But adding fractions is a solved problem, so that's not what you mean. So is a problem difficult only by virtue of being open? Like, FLT used to be difficult, but now that it's solved it's pretty much like adding fractions. Something for students to learn. But clearly that's not right either. Lot of thoughtful issues to sort out. Adding fractions is NOT difficult; and FLT still is. How to account for that? Aug 17, 2014 at 2:30
• ps -- "Closer to the boundary of knowledge." That's why FLT is still difficult even though it's solved. Aug 17, 2014 at 2:34
• so it's only difficult because no-one knew the answer not cos the answer was difficult to teach or learn? crazy.
– user6917
Aug 17, 2014 at 4:03
• The proof is very long and it requires lots of "machinery," a word mathematicians use to refer to complicated mathematical objects and the theory associated with these objects. For example, to answer some questions about the integer solutions of algebraic equations, you have to know a lot about complicated objects like schemes. Aug 17, 2014 at 4:33
• @user3293056 I'm making the point that FLT is still difficult even though we know the answer. Because it's still very close to the boundary of human knowledge; and because it takes many years of extreme specialization in the most advanced algebraic number theory to have any hope of working through the proof. Solvability isn't sufficient to make it not-difficult. It's still a very difficult problem, even though it's been solved. Aug 17, 2014 at 5:54

It's an interesting question to which I don't have a strongly philosophical answer to offer. What I can suggest is a concept I encountered elsewhere called a "wicked problem". I don't think wicked and difficult are perfectly synonymous here, but I would offer that all wicked problems are difficult problems.

Wikipedia will tell you:

1. The problem is not understood until after the formulation of a solution.
2. Wicked problems have no stopping rule.
3. Solutions to wicked problems are not right or wrong.
4. Every wicked problem is essentially novel and unique.
5. Every solution to a wicked problem is a 'one shot operation.'
6. Wicked problems have no given alternative solution

While I don't think that perfectly reflects what a difficult problem is, I think we could possibly derive a definition that a problem is difficult if it requires great effort to solve or nearly solve. A problem is subjectively difficult if it requires of the person in question great effort to solve or nearly solve. And a problem is objectively difficult if it requires specialist knowledge that is contemporary to solve or nearly solve.

But I have never seen difficult problems seen as a specifically philosophical problem.

It seems intuitively correct to you? That's strange. Take the subcase n = 3: There are no positive integers a, b, c such that a^3 + b^3 = c^3. Then modify it slightly: For which (positive or negative) integers k is it true that there are no positive integers a, b, and c such that a^3 + b^3 = c^3 + k? The case n = 3 of Fermat's Last Theorem is identical to this problem with k = 0.

It turns out that if k = 4 (modulo 9) or k = 5 (modulo 9) then it is very simple to show that there is no solution. For any other values of k except k = 0, it is believed that there will be an infinite number of solutions of which even one is often very, very, hard to find. But for the case k = 0, there is actually a very deep proof that there is not solution. I wouldn't call this in any way intuitive. It is actually quite counter-intuitive and surprising that FLT is true for n = 3. The same is true for n = 4, intuitively it seems reasonably likely that there would be solutions, but it turns out there is a proof that there are none. And after all, for n = 2 there is an infinite number of solutions. If n = 2 has an infinite number of solutions, what makes it intuitive to you that n = 3 would have none?

I seem to remember that the state of the art before Wiles' proof was that there are no solutions with n <= 125,000. It was indeed very, very unlikely at that point that solutions with n > 125,000 would be found, but by no means impossible. And by no means was it ever "intuitive" that this would be impossible.

• Very well put. It could also be said that some questions are difficult just because they are, for no other reason.
– nwr
Aug 23, 2014 at 23:34
• I'm generally sympathetic to this answer, but I'd hesitate to generalize from "The case \$n=3\$ is not intuitive" to "Cases with \$n>3\$ are not intuitive". The case \$n=3\$ is easier than, say, \$n=5\$, but it is arguably less intuitive, because there are simple heuristics suggesting that the answer should depend on whether \$3/n\$ is greater than 1, making \$n=3\$ the hairline case and making the truth of the theorem seem likelier as \$n\$ grows. (Of course "likely true for all higher \$n\$" is a much stronger statement than "likely true for any given higher \$n\$".) Mar 19, 2015 at 23:54
• i don't know why it seems intuitive for n=3, but it does.
– user64727
Feb 24 at 19:02

It's usually a process of attrition. The first question to ask is what makes this question interesting? It was a question asked by a famous mathematician — Fermat who died in 1665 and all of whose outstanding question apart from this one were solved by the early 1800s; then in 1816 the French Academy established a prize to solve this one; by then it was recognised that established techniques weren't enough; and it is this that makes the problem interesting — what are these new techniques? this turned out to be the development of new number systems and their geometrisation.

Mathematicians learn the art of asking the possible, and the important; sometimes it's difficult to discern where the boundary lies; that again comes with time; Fermat's last theorem is an obvious generalisation of Pythagora's theorem (a much more immediately important theorem — it proved its importance in antiquity, and then in the development of vector and metric manifolds important to both special and general relativity). Fermat probably believed it to be a possible question — it turned out to be an important one.

Again through the process of time, the percolation and maturing of these new techniques it's just maybe possible that Fermat's Last Theorem will be an exercise set for university students, rather than being one that exercised the best mathematical minds of the last three centuries. After all it took Galileo to start measuring just hw long it took a pendulum to swing, whereas any schoolboy interested in physics and bothered enough to do the experiment can actually do this.

It's worth quoting a comment by Cartier on this:

These two problems, Fermat and Riemann, are in some sense rather futile: Fermat’s problem concerned a very particular equation, and the Riemann hypothesis can be interpreted in terms of very subtle regularities in the apparently random distribution of prime numbers. In itself, a counterexample to the Riemann hypothesis, given the present state of our knowledge, would have very small “practical” consequences and would certainly not be a catastrophe.

A note on the solution: Fermat's last theorem is solved by an appeal to the Modularity theorem, which is a special case of what is currently called the Langland's program, and which appears to be the correct higher and non-abelian generalisation of class field theory - which basically tells you how badly factorisation fails for these new number systems (rings). It's this context that appears to be its natural home.